# ML Aggarwal Solutions for Chapter 15 Circles Class 10 Maths ICSE

Here, we are providing the solutions for Chapter 15 Circles from ML Aggarwal Textbook for Class 10 ICSE Mathematics. Solutions of the eleventh chapter has been provided in detail. This will help the students in understanding the chapter more clearly. Class 10 Chapter 15 Circles of ML Aggarwal Solutions for ICSE is one of the most important chapter for the board exams which is based on finding the radius, diameter of the circle, mid point of triangle, proving bisector, perpendicular and tangent to the circle, proving equilateral triangle and right angled triangle, finding the hypotenuse and perimeter of the triangle. We have also added chapter test and multiple choice questions.

### Exercise 15.1

1. Using the given information, find the value of x in each of the following figures:

(i)

(ii)
(iii)

(iv)
(v)

(vi)

(i) ∠ADB and ∠ACB are in the same segment.

∠ADB = ∠ACB = 50°

∠DAB + X + ∠ADB = 180°

= 42° + x + 50° = 180°

= 92° + x = 180°

x = 180° – 92°

x = 88°

(ii) In the given figure we have

= 32° + 45° + x = 180°

= 77° + x = 180°

x = 103°

(iii) From the given number we have

Because angles in the same segment

But ∠BAD = 20°

∠BCD = 20°

∠CEA = 90°

∠CED = 90°

Now in triangle CED,

∠CED + ∠BCD + ∠CDE = 180°

⇒ 90° + 20° + x = 180°

⇒ 110° + x = 180°

⇒ x = 180° – 110°

⇒ x = 70°

(iv) In ∆ABC

∠ABC + ∠ABC + ∠BAC = 180°

(Because sum of a triangle)

⇒ 69° + 31° + ∠BAC = 180°

⇒ ∠BAC = 180° – 100°

⇒ ∠BAC = 80°

Since, ∠BAC and ∠BAD are in the same

Segment.

∠BAD = x° = 80°

(v) Given ∠CPB = 120°, ∠ACP = 70°

To find, x° i,e., ∠PBD

Reflex∠CPB = ∠BPO + ∠CPA

⇒ 120° = ∠BPD + ∠BPD

(BPD = CPA are vertically opposite ∠s)

⇒ 2∠BPD = 120°

⇒ ∠PBD = 120°/2 = 60°

Also ∠ACP and PBD are in the same segment

∠PBD + ∠ACP = 70°

Now, In ∆PBD

∠PBD + ∠PDB + ∠BPD = 180°

(sum of all ∠s in a triangle)

70° + x° + 60° = 180°

⇒ x = 180° – 130°

⇒ x = 50°

(vi) ∠DAB = ∠BCD

(Angles in the same segment of the circle)

∠DAB = 25° (∠BCD = 25° given)

In ∆DAP,

Ex.∠CDA = ∠DAP + ∠DPA

⇒ x° = ∠DAB + ∠DPA

⇒ x° = 25° + 35°

⇒ x° = 60°

2. If O is the center of the circle, find the value of x in each of the following figures (using the given information):

(i)

(ii)

(iii)

(iv)

(v)
(vi)

(i) ∠ACB = ∠ADB

(Angles in the same segment of a circle)

But ∠ADB = x°

∠ABC = x°

Now in ∆ABC

∠CAB + ∠ABC + ∠ACB = 180°

⇒ 40° + 90° + x° = 180°

(AC is the diameter)

⇒ 130° +  = 180°

⇒ x° = 180° – 130° = 50°

(ii) ∠ACD = ∠ABD

(angles in the same segment)

∠ACD = x°

Now in triangle OAC,

OA = OC

(radii of the same circle)

∠ACO = ∠AOC

(opposite angles of equal sides)

Therefore, x° = 62°

(iii) ∠AOB + ∠AOC + ∠BOC = 360°

(sum of angles at a point)

∠AOB + 80° + 130° = 360°

⇒ ∠AOB + 210° = 360°

⇒ ∠AOB = 360° – 210° = 150°

Now arc AB subtends ∠AOB at the center ∠ACB at the remaining part of the circle

∠AOB = 2 ∠ACB

⇒ ∠ACB = ½ ∠AOB = ½ × 150° = 75°

(iv) ∠ACB + ∠CBD = 180°

⇒ ∠ABC + 75° = 180°

⇒ ∠ABC = 180° – 75° = 105°

Now arc AC Subtends reflex ∠AOC at the center and ∠ ABC at the remaining part of the circle.

Reflex∠AOC = 2 ∠ABC

= 2 × 105°

= 210°

(v) ∠AOC + ∠COB = 180°

⇒ 135° + ∠COB = 180°

⇒ ∠COB = 180° – 135° = 45°

Now arc BC Subtends reflex ∠COB at the center and ∠ CDB at the remaining part of the circle.

∠COB = 2 ∠CDB

⇒∠CDB = ½ ∠COB

= ½ × 45o = 45o/2 = 22 1/2o

(vi) Arc AB subtends ∠AOD at the center and ∠ACD at the remaining part of the Circle

∠AOD = 2 ∠ACB

⇒ ∠ACB = ½ ∠AOD = ½ × 70= 35o

∵ ∠CMO = 90o

∴ ∠AMC = 90o

(∠AMC + ∠CMO = 180o)

Now in ∆ACM

∠ACM + ∠AMC + ∠CAM = 180o

⇒ 35o + 90+ x= 180o

⇒ 125+ x= 180o

⇒ xo = 180 – 125= 55o

3. (a) In the figure (i) given below, AD || BC. If ACB = 35°. Find the measurement of DBC.

(b) In the figure (ii) given below, it is given that O is the center of the circle and AOC = 130°. Find ABC

(i)

(ii)

(a) Construction: Join AB

∠A = ∠C = 35° (Alt Angles)

⇒ ∠ABC = 35o

(b) ∠AOC + reflex ∠AOC = 360o

⇒ 130+ Reflex ∠AOC = 360o

⇒ Reflex ∠AOC = 360– 130= 230o

Now, arc BC Subtends reflex ∠AOC at the center and ∠ABC at the remaining part of the circle.

Reflex ∠AOC = 2 ∠ABC

⇒ ∠ABC =1/2 reflex ∠AOC

= ½ × 230= 115o

4. (a) In the figure (i) given below, calculate the values of x and y.

(b) In the figure (ii) given below, O is the center of the circle. Calculate the values of x and y.

(i)

(ii)

(a) ABCD is cyclic Quadrilateral

∠B + ∠D = 180°

⇒ y + 40° + 45= 180o

(y + 85= 180o)

⇒ y = 180– 85o = 95o

⇒ xo = 40

(b) Arc ADC Subtends ∠AOC at the center and
∠ ABC at the remaining part of the circle

∠AOC = 2 ∠ABC

⇒ x= 60o

Again, ABCD is a Cyclic quadrilateral

∠B + ∠D = 180o

(60+ y= 180o)

⇒ y = 180– 60= 120o

5. (a) In the figure (i) given below, M, A, B, N are points on a circle having center O. AN and MB cut at Y. If NYB = 50° and YNB = 20°, find MAN and the reflex angle MON.

(b) In the figure (ii) given below, O is the center of the circle. If AOB = 140° and OAC = 50°, find
(i)
ACB
(ii)
OBC
(iii)
OAB
(iv)
CBA

(i)

(ii)

(a) ∠NYB = 50°, ∠YNB = 20°.

In ∆YNB,

NYB + ∠YNB + YBN = 180o

⇒ 50+ 20+ ∠YBN = 180o

⇒ YBN + 70= 180o

⇒ YBN = 180– 70= 110o

But MAN = YBN

(Angles in the same segment)

⇒ MAN = 110o

Major arc MN subtend reflex MON at the

Centre and MAN at the remaining part of the choice.

Reflex MAN at the remaining part of the circle

Reflex MON = 2 MAN = 2×110=220o

(b) (i) AOB + reflex AOB = 360o

(Angles at the point)

⇒ 140+ reflex AOB = 360o

⇒ Reflex AOB = 360– 140= 220o

Now major arc AB subtends AOB+ OBC = 360o

⇒ 50+ 110+ 140OBC =360°

⇒ 300+ ∠OBC = 360°

⇒ ∠300o + ∠OBC = 360°

⇒ ∠OBC = 360o – 300o

⇒ ∠OBC = 60o

(ii) In Quadrilateral, OACB

∠OAC + ∠ACB + ∠AOB + ∠OBC = 360o

⇒ 50+ 110+ 140o + ∠OBC = 360o

⇒ 300o + ∠OBC = 360o

⇒ ∠OBC = 360o – 300o

⇒ ∠OBC =60o

(iii) In ∆OAB,

OA = OB

(Radii of the same circle)

⇒ ∠OAB + ∠OBA = 180o

⇒ 2∠OAB = 180o – 140= 40o

⇒ ∠OAB = 40o/2 = 20°

But, ∠OBC = 60o

∠CBA = ∠OBC – ∠OBA = 60o – 20= 40o

6. (a) In the figure (i) given below, O is the center of the circle and ∠PBA = 42°. Calculate the value of ∠PQB

(b) In the figure (ii) given below, AB is a diameter of the circle whose center is O. Given that ECD = EDC = 32°, calculate

(i) CEF

(ii) COF.

(i)

(ii)

In ∆APB,

∠APB = 90° (Angle in a semi-circle)

But ∠A + ∠APB + ∠ABP = 180° (Angles of a triangle)

⇒ ∠A + 90° + 42°= 180°

⇒ ∠A + 132° = 180°

⇒ ∠A = 180° – 132° = 48°

But ∠A = ∠PQB

(Angles in the same segment of a circle)

⇒ ∠PQB = 48o

(b) (i) in ∆EDC,

(Exterior angle of a triangle is equal to the sum of its interior opposite angels)

(ii) arc CF subtends ∠COF at the center and

∠CDF at the remaining part of the circle

∠COF = 2∠CDF = 2∠CDE

= 2×32= 2 ∠CDE

= 2×32

= 64

7.(a) In the figure (i) given below, AB is a diameter of the circle APBR. APQ and RBQ are straight lines, A = 35°, Q = 25°. Find (i) PRB (ii) PBR (iii) BPR

(b) In the figure (ii) given below, it is given that ABC = 40° and AD is a diameter of the circle. Calculate DAC.

(i)

(ii)

(a) (i) ∠PRB = ∠BAP

(Angles in the same segment of the circle)

∴ ∠PRB = 35° (∵ ∠BAP = 35° given)

(ii) In △PRQ,

Ext. ∠APR = ∠PRQ + ∠PQR

= ∠PRB + ∠Q

= 35°+25° = 60°

But ∠APB = 90° (Angle in a semi circle)

∴ ∠BPR = ∠APB - ∠APR

= 90° -60° = 30°

(iii) ∠APR = ∠ABR

(Angles in the same segment of the circle)

⇒ ∠ABR = 60°

In △PBQ,

Ext. ∠PBR = ∠Q + ∠BPQ

= 25° + 90° = 115°

(b) ∠B = ∠D (Angles in the same segment)

∴ ∠C = 40°

∠ACD = 90° (Angle in the semi circle)

∠ACD + ∠D + ∠DAC = 180° (Angles in a triangle)

⇒ 90° + 40° + ∠DAC = 180°

⇒ 130° + ∠DAC = 180°

⇒ ∠DAC = 180° – 130° = 50°

8. (a) In the figure given below, P and Q are centers of two circles intersecting at B and C. ACD is a straight line. Calculate the numerical value of x.

(b) In the figure given below, O is the circumcenter of triangle ABC in which AC = BC. Given that ACB = 56°, calculate

(i) CAB

(ii) OAC

Given that

(a) Arc AB subtends ∠APB at the center and ∠ACB at the remaining part of the circle

∠ACB = ½ ∠APB = ½ × 130= 65o

But ∠ACB + ∠BCD = 180(Linear Pair)

⇒ 65o + ∠BCD = 180o

⇒ ∠BCD = 180o – 65= 115o

Major arc BD subtends reflex ∠BQD at the

Centre and ∠BCD at the remaining part of the circle reflex ∠BQD = 2∠BCD

= 2×115o = 230°

But reflex ∠BQD + x = 360(Angles at a point)

⇒ 230+ x = 360o

⇒ x = 360– 230= 130o

(b) Join OC

In ∆ABC,

AC = BC

∠A = ∠B

But ∠A + ∠B + ∠C = 180o

⇒ ∠A + ∠A + 56° = 180°

⇒ 2∠A = 180– 56° = 124o

⇒ ∠A = 124/2 = 62or ∠CAB = 62°

OC is the radius of the circle. OC bisects ∠ACB.

∠OCA = ½∠ACB = ½×56= 28o

Now in ∆OAC

OA = OC (radii of the same Circle)

∠OAC = ∠OCA = 28o

### Exercise 15.2

1. If O is the center of the circle, find the value of x in each of the following figures (using the given information)

From the figure

(i) ABCD is a cyclic quadrilateral

Ext. ∠DCE = ∠BAD

Now arc BD subtends ∠BOD at the center

And ∠BAD at the remaining part of the circle.

∠BOD = 2∠BAD = 2 x

⇒ 2x = 150o (x = 75°)

(ii) ∠BCD + ∠DCE = 180(Linear pair)

⇒ ∠BCD + 80° = 180o

⇒ ∠BCD = 180° – 80° = 100o

Arc BAD subtends reflex ∠BOD at the

Centre and ∠BCD at the remaining part of the circle

Reflex ∠BOD = 2 ∠BCD

x= 2×100= 200o

(iii) In ∆ACB,

∠CAB + ∠ABC + ∠ACB = 180(Angles of a triangle)

But ∠ACB = 90(Angles of a semicircle)

25+ 90+ ∠ABC = 180o

⇒ 115+ ∠ABC = 180o

⇒ ∠ABC = 180– 115° =65o

ABCD is a cyclic quadrilateral

∠ABC + ∠ADC = 180(Opposite angles of a cyclic quadrilateral)

⇒ 65+ x=180o

⇒ xo = 180o - 65o = 115o

2. (a) In the figure (i) given below, O is the center of the circle. If ∠AOC = 150°, find (i) ∠ABC (ii) ∠ADC

(b) In the figure (i) given below, AC is a diameter of the given circle and BCD = 75°. Calculate the size of (i) ABC (ii) EAF.

(a) Given, ∠AOC = 150° and AD = CD

We know that an angle subtends by an arc of a circle at the center is twice the angle subtended by the same arc at any point on the remaining part of the circle.

(i) ∠AOC = 2×∠ABC

∠ABC = ∠AOC/2 = 150o/2 = 75o

(ii) From the figure, ABCD is a cyclic quadrilateral

∠ABC + ∠ADC = 180(Sum of opposite angels in a cyclic quadrilateral is 180o)

⇒ 75o + ∠ADC = 180o

⇒ ∠ADC + 180– 75o

⇒ ∠ADC = 105o

(b) (i) AC is the diameter of the circle

∠ABC = 90(Angle in a semi-circle)

(ii) ABCD is a cyclic quadrilateral

∠BAD + ∠BCD = 180o

⇒ ∠BAD + 75= 180(∠BCD = 75o)

⇒ ∠BAD = 180- 75o = 105o

But ∠EAF = ∠BAD (Vertically opposite angles)

∠EAF = 105o

3. (a) In the figure, (i) given below, if DBC = 58° and BD is a diameter of the circle, calculate

(i) BDC (ii) BEC (iii) BAC

(b) In the figure (if) given below, AB is parallel to DC, BCE = 80° and BAC = 25°. Find:

(a) ∠DBC = 58°

BD is diameter

∠DCB = 90° (Angle in semi-circle)

(i) In ∆BDC

∠BDC + ∠DCB + ∠CBD = 180°

⇒ ∠BDC = 180°- 90° – 58° = 32°

(ii) BEC = 180o – 32= 148(opposite angles of cyclic quadrilateral)

(iii) ∠BAC = ∠BDC = 32(Angles in same segment)

(b) in the figure, AB ∥DC

∠BCE = 80o and ∠BAC = 25o

ABCD is a cyclic Quadrilateral and DC is

Production to E

(i) Ext, ∠BCE = interior ∠A

⇒ 80o = ∠BAC + ∠CAD

⇒ 80o = 25o + ∠CAD

⇒ ∠CAD = 80o – 25o = 55o

(ii) But ∠CAD = ∠CBD (Alternate angels)

∠CBD = 55o

(iii) ∠BAC = ∠BDC (Angles in the same segments)

∠BDC = 25(∠BAC = 25o)

Now AB ∥ DC and BD is the transversal

∠BDC = ∠ABD

⇒ ∠ABD = 25o

⇒ ∠ABC = ∠ABD + ∠CBD = 25+ 55o = 80o

But ∠ABC + ∠ADC = 180(opposite angles of a cyclic quadrilateral)

⇒ 80o + ∠ADC = 180o

⇒ ∠ADC = 180– 80o = 100o

4. (a) In the figure given below, ABCD is a cyclic quadrilateral. If ADC = 80° and ACD = 52°, find the values of ABC and CBD.

(b) In the figure given below, O is the center of the circle. AOE =150°, DAO = 51°. Calculate the sizes of BEC and EBC.

(a) In the given figure, ABCD is a cyclic quadrilateral

∠ADC = 80° and ∠ACD = 52°

To find the measure of ∠ABC and ∠CBD

ABCD is a Cyclic Quadrilateral

∠ABC + ∠ADC = 180(Sum of opposite angles = 180o)
⇒ ∠ABC + 80= 180o

∠AOE = 150o, ∠DAO = 51o

To find ∠BEC and ∠EBC

ABED is a cyclic quadrilateral

Ext. ∠BEC = ∠DAB = 51o

⇒ ∠AOE = 150o

⇒ Ref. ∠AOE = 360– 150o = 51o

⇒ ∠AOE = 150o

⇒ Ref. ∠AOE = 360o – 150o = 210o

Now, arc. ABE subtends ∠AOE at the Centre

And ∠ADE at the remaining part of the circle.

⇒ ∠ADE = ½ ref ∠AOE = ½ ×210= 105o

But Ext. ∠EBC = ∠ADE = 105o

Hence, ∠BEC = 51and ∠EBC = 105o

(b) In the given figure, O is the centre of the circle.

∠AOE = 150°, ∠DAO = 51°

To find ∠BEC and ∠EBC

ABED is a cyclic quadrilateral

∴ Ext. ∠BEC = ∠DAB = 51°

∵ ∠AOE = 150°

∴ Ref. ∠AOE = 360° - 150° = 210°

Now arc ABE subtends ∠AOE at the centre and ∠AE at the remaining part of the circle.

∴ ∠ADE = ½ Ref. ∠AOE = 1/2 ×210° = 105°

But, Ext. ∠EBC = ∠ADE = 105°

Hence, ∠BEC = 51° and ∠EBC = 105°

5. (a) In the figure (i) given below, ABCD is a parallelogram. A circle passes through A and D and cuts AB at E and DC at F. Given that BEF = 80°, find ABC.

(b) In the figure (ii) given below, ABCD is a cyclic trapezium in which AD is parallel to BC and B = 70°, find:

(i) BAD (ii) DBCD.

(i)

(ii)

Ext. ∠FEB = ∠ADF

⇒ ∠ADF = 80°

ABCD is a parallelogram

∠B = ∠D = ∠ADF = 80°

or ∠ABC = 80°

(b) In trapezium ABCD, AD || BC

(i) ∠B + ∠A = 180°

⇒ 70° + ∠A = 180°

⇒ ∠A = 180° – 70° = 110°

(ii) ABCD is a cyclic quadrilateral

∠A + ∠C = 180°

⇒ 110° + ∠C = 180°

⇒ ∠C = 180° – 110° = 70°

Therefore, ∠BCD = 70°

6. (a) In the figure given below, O is the center of the circle. If BAD = 30°, find the values of p, q and r.

(a) In the figure given below, two circles intersect at points P and Q. If A = 80° and D = 84°, calculate

(i) QBC

(ii) BCP

(i) ABCD is a cyclic quadrilateral

∠A + ∠C = 180o

⇒ 30+ p = 180o

⇒ p = 180o – 30= 150o

(ii) Arc BD subtends ∠BOD at the center

And ∠BAD at the remaining part of the circle

⇒ q = 2×30= 60o

∠BAD = ∠BED are in the same segment of the circle

⇒ ∠BAD = ∠BED

⇒ 30= r

⇒ r = 30o

Join PQ,

AQPD is a cyclic quadrilateral

∠A + ∠QPD = 180o

(b) Join PQ

AQPD is a cyclic quadrilateral

A + QPD = 180o

80° + QPD = 180°

QPD = 180° - 80° = 100°

And D + AQP = 180°

84° + AQP = 180°

AQP = 180° - 84° = 96°

Now, PQBC is a cyclic quadrilateral,

Ext. QPD = QBC

QBC = 100°

And ext. AQP = BCP

BCP = 96°

7. (a) In the figure given below, PQ is a diameter. Chord SR is parallel to PQ. Given PQR = 58°, calculate (i) RPQ (ii) STP

(T is a point on the minor arc SP)

(b) In the figure given below, if ACE = 43° and CAF = 62°, find the values of a, b and c.

(a) In ∆PQR,

∠PRQ = 90° (Angle in a semi-circle) and ∠PQR = 58°

∠RPQ = 90° – ∠PQR = 90° – 58° = 32°

SR || PQ (given)

∠SRP = ∠RPQ = 32(Alternate angles)

Now, PRST is a cyclic quadrilateral,

∠STP + ∠SRP = 180o

⇒ ∠STP = 180– 32o = 148o

(b) In the given figure,

∠ACE 43and ∠CAF = 620

Now, in ∆AEC

∠ACE + ∠CAE + ∠AEC = 180o

⇒ 43+ 62+ ∠AEC = 180o

⇒ 105+ ∠AEC = 180o

⇒ ∠AEC = 180– 105° = 75o

But ∠ABD + ∠AED = 180° (sum of opposite angles of acyclic quadrilateral) and,

∠AED = ∠AEC

⇒ a + 75= 180o

⇒ a = 180– 75– 105o

but ∠EDF = ∠BAE (Angles in the alternate segment)

8. (a) In the figure (i) given below, AB is a diameter of the circle. If ADC = 120°, find CAB.

(b) In the figure (ii) given below, sides AB and DC of a cyclic quadrilateral ABCD are produced to meet at E, the sides AD and BC are produced to meet at F. If x : y : z = 3 : 4 : 5, find the values of x, y and z.

(a) Construction: Join BC, and AC then

ABCD is a cyclic quadrilateral.

Now in ∆DCF

Ext. ∠2 = x + z and,

In ∆CBE,

Ext. ∠1 = x + y

Adding (i) and (ii)

x + y + x + z = ∠1 + ∠2

2 x + y + z = 180(ABCD is a cyclic quadrilateral)

But x : y : z = 3 : 4 : 5

x/y = ¾ (y = 4/3 x)

x/z = 3/5 (z = 5/3)

### Exercise 15.3

1. Find the length of the tangent drawn to a circle of radius 3cm, from a point distinct 5 cm from the center.

In a circle with center O and radius 3cm and p is at a distance of 5 cm.

OT = 3 cm, OP = 5 cm

OT is the radius of the circle

OT ⊥ PT

Now, in right ∆ OTP.

By Pythagoras axiom,

OP= OT+ PT2

⇒ (5)= (3)+ PT2

⇒ PT= (5)2 – (3)2 = 25 – 9 = 16 = (4)2

⇒ PT = 4 cm.

2A point P is at a distance 13 cm from the center C of a circle and PT is a tangent to the given circle. If PT = 12 cm, find the radius of the circle.

CT is the radius

CP = 13 cm and tangent PT = 12 cm

CT is the radius and TP is the tangent

CT is perpendicular TP

Now in right angled triangle CPT,

CP2 = CT2 + PT2 [using Pythagoras axiom]

⇒ (13)2 = (CT)2 + (12)2

⇒ 169 = (CT)2 + 144

⇒ (CT)= 169 -144 =25 = (5)2

⇒ CT = 5 cm.

Hence, the radius of the circle is 5 cm.

3. The tangent to a circle of radius 6 cm from an external point P, is of length 8 cm. Calculate the distance of P from the nearest point of the circle.

Radius of the circle = 6 cm

and length of tangent = 8 cm

Let OP be the distance

i.e. OA = 6 cm, AP = 8 cm,

OA is the radius

OA ⊥ AP

Now In right ∆OAP,

OP= OA+ AP(By Pythagoras axiom)

= (6)+ (8)2

= 36 + 64

= 100

= (10)2

⇒ OP = 10 cm.

4. Two concentric circles are of the radii 13 cm and 5 cm. Find the length of the chord of the outer circle which touches the inner circle.

Two concentric circles with center O

OP and OB are the radii of the circles respectively, then

OP = 5 cm, OB = 13 cm.

AB is the chord of outer circle which touches the inner circle at P.

OP is the radius and APB is the tangent to the inner circle.

In the right angled triangle OPB, by Pythagoras axiom,

OB2 = OP2 + PB2

⇒ 132 = 52 + PB2

⇒ 169 = 25 + PB2

⇒ PB2 = 169 – 25 = 144

⇒ PB = 12 cm

But P is the mid-point of AB.

AB = 2PB = 24 cm

5. Two circles of radii 5 cm and 2-8 cm touch each other. Find the distance between their centers if they touch :

(i) externally

(ii) internally.

Radii of the circles are 5 cm and 2.8 cm.

i.e. OP = 5 cm and CP = 2.8 cm.

(i) When the circles touch externally, then the distance between their centers = OC = 5 + 2.8 = 7.8 cm.

(ii) When the circles touch internally, then the distance between their centers = OC = 5.0 – 2.8 = 2.2 cm

6. (a) In figure (i) given below, triangle ABC is circumscribed, find x.

(b) In figure (ii) given below, quadrilateral ABCD is circumscribed, find x.

(a) From A, AP and AQ are the tangents to the circle

∴ AQ = AP = 4cm

But AC = 12 cm

CQ = 12 – 4 = 8 cm.

From B, BP and BR are the tangents to the circle

BR = BP = 6 cm.

Similarly, from C,

CQ and CR the tangents

CR = CQ = 8 cm

⇒ x = BC = BR + CR = 6 cm + 8 cm = 14 cm

(b) From C, CR and CS are the tangents to the circle.

CS = CR = 3 cm.

But, BC = 7 cm.

⇒ BS = BC – CS = 7 – 3 =4 cm.

Now from B,BP and BS are the tangents to the circle.

BP = BS = 4 cm

From A, AP and AQ are the tangents to the circle.

AP = AQ = 5cm

⇒ x = AB = AP + BP = 5 + 4 = 9 cm

7. (a) In figure (i) given below, quadrilateral ABCD is circumscribed; find the perimeter of quadrilateral ABCD

(b) In figure (ii) given below, quadrilateral ABCD is circumscribed and AD DC ; find x if radius of the circle is 10 cm.

(a) From A, AP and AS are the tangents to the circle

∴ AS = AP = 6

From B, BP and BQ are the tangents

∴ BQ = BP = 5

From C, CQ and CR are the tangents

CR = CQ

From D, DS and DR are the tangents

DS = DR = 4

Therefore, perimeter of the quadrilateral ABCD

= 6 + 5 + 5 + 3 + 3 + 4 + 4 + 6

= 36 cm

(b) in the circle with center O, radius OS = 10 cm

PB = 27 cm, BC = 38 cm

OS id the radius and AD is the tangent.

Therefore, OS perpendicular to AD.

SD = OS = 10 cm.

Now from D, DR and DS are the tangents

To the circle

DR = DS = 10 cm

From B, BP and BQ are tangents to the circle.

BQ = BP = 27 cm.

⇒ CQ = CB – BQ = 38 – 27 = 11 cm.

Now from C, CQ and CR are the tangents to the circle

CR = CQ = 11 cm.

⇒ DC = x = DR + CR = 10 + 11 = 21 cm

8. (a) In the figure (i) given below, O is the center of the circle and AB is a tangent at B. If AB = 15 cm and AC = 7.5 cm, find the radius of the circle.

(b) In the figure (ii) given below, from an external point P, tangents PA and PB are drawn to a circle. CE is a tangent to the circle at D. If AP = 15 cm, find the perimeter of the triangle PEC.

(i) Join OB

∠OBA = 90° (Radius through the point of contact is perpendicular to the tangent)

OB= OA– AB2

⇒ r= (r + 7.5)– 152

⇒ r= r+ 56.25 + 15r – 225

⇒ 15r = 168.75

⇒ r = 11.25

Hence, radius of the circles = 11.25 cm

(ii) In the figure, PA and PB are the tangents

Drawn from P to the circle.

CE is tangent at D

AP = 15 cm

PA and PB are tangents to the circle

AP = BP = 15 cm

Similarly EA and ED are tangents

EA = ED

Similarly BC = CD

Now perimeter of triangle PEC,

= PE + EC + PC

= PE + ED + CD + PC

= PE + EA + CB + PC (ED = EA and CB = CD)

= AP + PB

= 15 + 15

= 30 cm.

9. (a) If a, b, c are the sides of a right triangle where c is the hypotenuse, prove that the radius r of the circle which touches the sides of the triangle is given by

r = (a+b–c)/2

(b) In the given figure, PB is a tangent to a circle with center O at B. AB is a chord of length 24 cm at a distance of 5 cm from the center. If the length of the tangent is 20 cm, find the length of OP.

(a) Let the circle touch the sides BC, CA and AB of the right triangle ABC at points D, E and F respectively, where BC = a, CA = b and AB = c (as showing in the given figure).

As the lengths of tangents drawn from an external point to a circle are equal

AE = AF, BD = BF and CD = DE

OD ⊥ BC and OE ⊥ CA (tangents is ⊥ to radius)

ODCE is a square of side r

DC = CE = r

AF = AE = AC – EC = b – r and,

BF = BD = BC – DC = a – r

Now,

AB = AF + BF

c = (b – r) + (a – r)

⇒ 2r = a + b – c

⇒ r = (a+b–c)/2

⇒ OP= 400 + 169

⇒ OP = a − √569cm

10. Three circles of radii 2 cm, 3 cm and 4 cm touch each other externally. Find the perimeter of the triangle obtained on joining the centers of these circles.

Three circles with centers A, B and C touch each other externally at P, Q and R respectively and the radii of these circles are 2 cm, 3 cm and 4 cm.

By joining the centers of triangle ABC formed in which,

AB = 2 + 3 = 5 cm

BC = 3 + 4 = 7 cm

CA = 4 + 2 = 6 cm

Therefore, perimeter of the triangle ABC = AB + BC + CA

= 5 + 7 + 6

= 18 cm

11. (a) In the figure (i) given below, the sides of the quadrilateral touch the circle. Prove that AB + CD = BC + DA.

(b) In the figure (ii) given below, ABC is triangle with AB = 10 cm, BC = 8 cm and AC = 6 cm (not drawn to scale). Three circles are drawn touching each other with vertices A, B and C as their centres. Find the radii of the three circles.

(a) Given: Sides of quadrilateral ABCD touch the circle at P, Q, R and S respectively.

To prove: AB + CD = BC + DA

Proof:

Tangents from A, AP and AS are to the circle.

∴ AP = AS

Similarly,

PB = BQ, CR = CQ and DR = DS

Adding them, we get

AP + PB + CR + DR = AS + BQ + CQ + DS

⇒ AP + PB + CR + DR = BQ + CQ + AS + SD

⇒ AB + CD = BC + DA

(b) AB = 10 cm, BC = 8 cm, AC = 6 cm

BP + PA = 10 cm …(i)

BC = 8 cm (given)

BR + RC = 8 cm

∴ BP + RC = 8 …(ii)

Again, AC = 6 cm

AQ + QC = 6 cm

PA + RC = 6 cm ...(iii)

Adding, (i) + (ii) + (iii) we have

2 (BP + PA + RC) = 24 cm

∴ BP + PA + RC = 12 cm …(iv)

Now, (iv) – (i) we have

RC = 12 - 10 = 2 cm

Also, (iv) – (ii) we have

PA = 12 – 8 = 4 cm

Also, (iv) – (iii) we have

BP = 12 – 6 = 6 cm

Radii: BP = 6 cm, PA = 4 cm, RC = 2 cm

12. (a) In the figure (i) PQ = 24 cm, QR = 7 cm and PQR = 90ͦ Find the radius of the inscribed circle Î”PQR.

(b) In the figure (ii) given below, two concentric circles with centre O are of radii 5 cm and 3 cm. From an external point P, tangents PA and PB are drawn to the circles. If AP = 12 cm, find BP.

(a) In the figure, a circle is inscribed in the triangle PQR which touches the sides. O is entre of the circle

PQ = 24 cm, QR = 7 cm and ∠PQR = 90°

OM is joined. Join OL.

∵ OL and OM are the radii of the circle

∴ OL ⊥ PQ and OM ⊥ BC and ∠PQR = 90°

∴ QLOM is a square

∴ OL = OM = QL = QM.

Now in Î”PQR ∠O = 90°

∴ PQ2 = PQ2 + QR2 (Pythagoras theorem)

= (24)2 + (7)2

= 576 + 49

= 625

= (25)2

∴ P = 25 cm

Let OB = x, then

QM = QL = x

∵ PL and PN are the tangents to the circle.

PL = PN

⇒ PQ – LQ = PR – RN

⇒ 24 – x = 25 – PM (∵ RN = RM)

⇒ 24 – x = 25 – (QR – QM)

⇒ 24 – x = 25 – (7 – x)

⇒ 24 – x = 25 – 7 + x

⇒ 24 – 25 + 7 = 2x

⇒ 2x = 6

⇒ x = 6/2 = 3 cm

∴ Radius of the incircle = 3 cm

(b) Radius of outer circle = 5 cm

And radius of inner circle = 3 cm

∴ OA = 5 cm

OB = 3 cm and AP = 12 cm

∵ OA = 5 cm

OB = 3 cm and AP = 12 cm

∵ OA = 5 cm

OB = 3 cm and AP = 12 cm

∵ OA is radius and AP is the tangent.

∴ OA ⊥ AP

∴ In right Î”OAP

OP2 = OA2 + AP2

= (5)2 + (12)2

= 25 + 144

= 169

= (13)2

∴ OP = 13

Similarly in right ∠OBP

OP2 = OB2 + BP2

⇒ (13)2 = (3)2 + BP2

⇒ 169 = 9 + BP2

⇒ BP2 = 169 – 9 = 160

13. (a) In the figure (i) given below, AB = 8 cm and M is mid-point of AB. Semi-circles are drawn on AB, AM and MB as diametres. A circle with centre C touches all three semi-circles as shown, find its radius.

(b) In the figure (ii) given below, equal circles with centres O and O’ touch each other at X. OO’ is produced to meet a circle O’ at A. AC is tangent to the circle whose centre is O. O’D is perpendicular to AC. Find the value of :

(i) AO'/AO

(ii) (area of Î”ADO')/(area of Î”ACO)

(a) Let x be the radius of the circle with centre C and radii of each equal

Semi circles = 4/2 = 2 cm

CP = x + 2 and CM = 4 – x

∴ In right Î”PCM

(PC)2 = (PM)2 + (CM)2

⇒ (x + 2)2 = (2)2 + (4 – x)2

⇒ x2 + 4x + 4 = 4 + 16 – 8x + x2

⇒ 4x + 8x + 4 – 4 – 16 = 0

⇒ 12x = 16

⇒ x = 16/12 = 4/3 cm

∴ Radius = 4/3 cm.

(b) Two equal circles with centre O and O’ touch each other externally at X. OO' is joined and produced to meet the circle at A. AC is the tangent to the circle with centre O. O'D is ⊥ AC. Join AC.

∵ OC is radius and AC is tangent, then OC ⊥ AC.

Let radius of each equal circle = r.

In Î”ADO' and Î”ACO.

∠A = ∠A (Common)

∠D = ∠C (each 90°)

∴ Î”ADO' ~ Î”ACO (AA axiom of similarity)

(i) ∴ AO'/AO = r/3r = 1/3

(ii) ∵ Î”ADO' ~ Î”ACO

∴ (area of Î”ADO')/(area of Î”ACO)

= (AO')2/(AO)2

= (1/3)2 (from (i))

= 1/9

14. The length of the direct common tangent to two circles of radii 12 cm and 4 cm is 15 cm. Calculate the distance between their centres.

Let R and r be the radii of the circles with centre A and B respectively

Let TT’ be their common tangent.

∴ (TT)2 = (AB)2 – (R – r)2

⇒ (15)2 = (AB)2 – (12 – 4)2

⇒ 225 = AB2 – (8)2

⇒ (AB)2 = 225 + 64

= 289

= (17)2

∴ AB = 17 cm

Hence, distance between their centres = 17 cm.

15. Calculate the length of a direct common tangent to two circles of radii 3 cm and 8 cm with their centres 13 cm apart.

Let A and B be the centres of the circles whose radii are 8 cm and 3 cm and

Let TT’ length of their common tangent and AB = 13 cm.

Now (TT’)2 = (AB)2 – (R – r)2

= (13) - (8 – 3)2

= 169 – (5)2

= 169 – 25

= 144

= (12)2

∴ TT’ = 12 cm

Hence, length of common tangent = 12 cm.

16. In the given figure, AC is a transverse common tangent to two circles with centres P and Q and of radii 6 cm and 3 cm respectively. Given that AB = 8 cm, calculate PQ.

AC is a transverse common tangent to the two circles with centre P and Q and of radii 6 cm and 3 cm respectively

AB = 8 cm. Join AP and CQ.

In right Î”PAB,

PB2 = PA2 + AB2

= 62 + 82

= 36 + 64

= 100

= 102

∴ PB = 10 cm

Now in Î”PAB and Î”BCQ,

∴ ∠A = ∠C (each 90°)

∠ABP = ∠CBQ (vertically opposite angles)

∴ Î”PAB ~ ∠BCQ (AA axiom of similarity)

∴ AP/CQ = PB/BQ

⇒ 6/3 = 10/BQ

⇒ BQ = (10×3)/6 = 5 cm

∴ PQ = PB + BQ

= 10 + 5

= 15 cm.

17. Two circles with centres A, B are of radii 6 cm and 3 cm respectively. If AB = 15 cm, find the length of a transverse common tangent tangent to these circles

AB = 15 cm.

Radius of the circle with centre A = 6 cm

And radius of second circle with radius B = 3 cm.

Let AP = x, then PB = 15 – x

In Î”ATP and Î”SBP

∠T = ∠S (each 90°)

∠APT = ∠BPS (vertically opposite angles)

∴ Î”ATP ~ Î”SBP (AA axiom of similarity)

AT/BS = AP/PB

⇒ 6/3 = x/(15–x)

⇒ x/(15–x) = 6/3

⇒ 3x = 90 – 6x

⇒ 3x + 6x = 90

⇒ 9x = 90

∴ x = 90/9 = 10

∴ AP = 10 cm,

PB = 15 – 10 = 5 cm

Now in right Î”ATP,

AP2 = AT2 + TP2

⇒ (10)2 = (6)2 + TP2

⇒ 100 = 36 + TP2

⇒ TP2 = 100 – 36 = 64

⇒ TP2 = (8)2

∴ TP = 8 cm

Similarly in right Î”PSB,

PB2 = BS2 + PS2

⇒ (5)2 = (3)2 + PS2

⇒ 25 = 9 + PS2

⇒ PS2 = 25 – 9

= 16

= (4)2

∴ PS = 4 cm

Hence TS = TP + PS

= 8 + 4

= 12 cm

18. (a) In the figure (i) given below, PA and PB are tangents at a points A and B respectively of a circle with centre O. Q and R are points on the circle. If APB = 70°, find

(i) AOB

(ii) AQB

(iii) ARB

(b) In the figure (ii) given below, two circles touch internally at P from an external point Q on the common tangent at P, two tangents QA and QB are drawn to the two circles. Prove that QA = QB.

(a) To find: (i) ∠AOB, (ii) ∠AQB, (iii) ∠ARB

Given : PA and PB are tangents at the points A and B respectively of a circle with centre O and OA and OB are radii on it.

∠APB = 70°

Construction: Join AB

∴ OAB = ∠OBA

(i) Since, AP and BP are tangents to the circle and OA and OB are radii on it.

∴ OA ⊥ OP and OB ⊥ BP

∴ ∠AQB = 180° - 70°

= 110°

(ii) Arc AB subtends ∠AOB at the centre and ∠AOB at the remaining part of the circle.

∴ ∠AQB = 2∠AOB

⇒ ∠AQB = ½ ∠AOB

⇒ ∠AQB = 1/2∠AOB

⇒ ∠AQB = ½ × 110° = 55°

(iii) Now in Î”OAB,

OA = OB

∴ ∠OAB = ∠OBA (Radii of the same circle)

= ½ (180° - 110°)

= ½ × 70°

= 35°

We know, ∠AOB = 110°

Reflex ∠AOB = 360° - 110°

= 250°

∴ ∠ARB = ½ of reflex ∠AOB (∵ Angle at the centre of a circle = double the angle at the remaining part of the circle)

= ½ × 250°

= 120°

(b) In the fig. (ii)

Two circles touch each other internally at P. From a point Q, outside, common tangents QP, QA and QB are drawn to the two circles.

To prove : QA = QB

Proof : From Q, QA and QP are the tangents to the outer circle.

∴ QP = QA ...(i)

Similarly, from Q, QB and QP are the tangents to the inner circle.

∴ QP = QB …(ii)

From (i) and (ii)

QA = QB

Hence, proved.

19. In the given figure, AD is a diameter of a circle with centre O and AB is tangent at A. C is is a point on the circle such that DC produced intersects the tangent of B. If ABC = 50°, find AOC.

Given AB is tangent to the circle at A and OA is radius, OA ⊥ AB

In Î”ABD

∠OAB + 90° + ∠ADC = 180°

⇒ ∠OAB + 90° + 50° = 180°

⇒ ∠OAB + 140° = 180°

⇒ ∠OAB + 180° - 140° = 40°

(∵ Angle at centre of a circle = double the angle at the remaining part of circle)

∴ ∠ABC = 2 × 40°

= 80°

20. In the given figure, tangents PQ and PR are drawn from an external point P to a circle such that RPQ = 30°. A chord RS is drawn parallel to the tangent PQ, Find RQS

In the given figure,

PQ and PR are tangents to the circle with centre O drawn from P

∠RPQ = 30°

Chord RS || PQ is drawn

To find ∠RQS

∴ PQ = PR (tangents to the circle)

∴ ∠PRQ = ∠PQR

But ∠RPO = 30°

∴ ∠PRO = ∠PQR = (180° - 30°)/2

= 150°/2

= 75°

And ∠RSQ = ∠QRS (∵ QR = QS)

= 75°

Now in Î”QRS,

∠RQS = 180° - (∠RSQ + ∠QRS)

= 180° - (75° + 75°)

= 180° - 150°

= 30°

21. (a) In the figure (i) given below, PQ is a tangent to the circle at A, DB is a diameter,

ADB = 30° and CBD = 60°, calculate (i) QAB (ii) PAD (iii) CDB.

(b) In the figure (ii) given below, ABCD is a cyclic quadrilateral. The tangent to the circle at B meets DC produced at F. If EAB = 85° and BFC = 50°, find CAB.

(a) PQ is tangent and Ad is chord

(i) ∴ ∠QAB = ∠BDA = 30° (Angles in the alternate segment)

∠DAB = 90° (Angle in a semi-circle)

And ∠ADB = 30°

∴ ∠ABD = 180° - (90° + 30°)

= 180° - 120°

= 60°

But ∠PAD = ∠ABD (Angles in the alternate segment)

= 60°

(iii) In right Î”BCD

∠BCD = 90° (Angle is a semi-circle)

∠CBD = 60° (given)

∴ ∠CBD = 180° - (90° + 60°)

= 180° - 150°

= 30°

(b) ABCD is a cyclic quadrilateral.

∴ Ext. ∠EAB = ∠BCD

∴ ∠BCD = 85°

But ∠BCD + ∠BCF = 180° (Linear pair)

⇒ 85° + ∠BCF = 180°

∠BCF = 180° - 85° = 95°

Now in Î”BCF,

∠BCF + ∠BFC + ∠CBF = 180°

⇒ 95° + 50° + ∠CBF = 180°

⇒ 145° + ∠CBF = 180°

⇒ ∠CBF = 180° – 145° = 35°

∵ BF is a tangent and BC is the chord.

∴ ∠CAB = ∠CBF (Angles in the alternate segment)

⇒ ∠CAB = 35°

22. (a) In the figure (i) given below, O is the centre of the circle and SP is a tangent. If SRT = 65°, find the value of x, y and z.

(b) In the figure (ii) given below, O is the centre of the circle. PS and PT are tangents and SPT = 84°, Calculate the sizes of the angles of the angles TOS and TQS.

Consider the following figure:

TS ⊥ SP,

∠TSR = ∠OSP = 90°

In Î”TSR,

∠TSR + ∠TRS + ∠RTS = 180°

⇒ 90° + 65° + x = 180°

⇒ x = 180° - 90° - 65°

⇒ x = 25°

⇒ y = 2x

[Angle subtended at the centre is double that of the angle subtended by the arc at the same centre]

⇒ y = 2 × 25°

⇒ y = 50°

In Î”OSP,

∠OSP + ∠SPO + ∠POS = 180°

⇒ 90° + z + 50° = 180°

⇒ z = 180° - 140°

∴ z = 40°

Hence x = 25°, y = 50° and z = 40°

(b) ∵ TP and SP are the tangents and OS and OT are the radii

∴ OS ⊥ SP and OT ⊥ SP

∴ ∠OSP = ∠OTP = 90°

∠SPT = 84°

But ∠SOT + ∠OTP + ∠TPS + OSP = 360°

⇒ ∠SOT + 90° + 84° + 90° = 360°

⇒ ∠SOT + 264° = 360°

⇒ ∠SOT = 360° - 26°

⇒ ∠SOT = 96°

And reflex ∠SOT = 360° – 96° = 264°

Now major arc ST subtends reflex ∠SOT at the centre and ∠TQS at the remaining part of the circle.

∴ ∠TQS = ½ reflex

∠SOT = ½ × 264° = 132°

Hence ∠TOS = 96° and ∠TQS = 132°

23. In the adjoining figure, O is the centre of the circle. Tangents to the circle at A and B meet at C. If ACO = 30°, find

(i) BCO

(ii) AOR

(iii) APB

(i) ∠BCO = ∠ACO = 30°

(∵ C is the intersecting point of tangent AC and BC)

(ii) ∠OAC = ∠OBC = 90°

∵ ∠AOC = ∠BOC = 180° - (90°+ 30°) = 60°

(∵ ∠AOB = ∠AOC + ∠BOC)

= 60° + 60°

= 120°

(iii) ∠APB = ½ ∠AOB

= 120°/2

= 60°  [∵ angle subtended at the remaining part of the circle is half the subtended at the centre]

24. (a) In the figure (i) given below, O is the centre of the circle. The tangent at B and D meet at P. If AB is parallel to CD and ABC = 55°. Find:

(i) BOD (ii) BPD

(b) In the figure (ii) given below. O is the centre of the circle. AB is a diameter, TPT’ is a tangent to the circle at P. If BPT’ = 30°, calculate:

(i) APT (ii) BOP.

(a) AB || CD

(i) ∠ABC = ∠BCD (Alternate angles)

⇒ ∠BCD = 55°

Now arc BD subtends ∠BOD at the centre and ∠BCD at the remaining part of the circle.

∴ ∠BOD = 2 ∠BCD

= 2 × 55°

= 110°

OB is radius and BP is tangent

∴ OB ⊥ BP°

Similarly, OD ⊥ DP

Now in quad. OBPD,

∠BOD + ∠ODP + ∠OBP + ∠BPD = 360° (Angles of quadrilateral)

⇒ 110° + 90° + 90° + ∠BPD = 360°

⇒ 290° + ∠BPD = 360°

⇒ ∠BPD = 360° - 290° = 70°

(b) TPT’ is the tangent.

(i) AB is diameter.

∴ ∠APB = 90° (Angle in a semi-circle)

And ∠BPT’ = 30°

∴ ∠APT = 180 ̊ - (90° + 30°)

= 180° – 120°

= 60°

(ii) TPT’ is tangent and PB is the chord.

∴ ∠BAP = ∠BPT' (Angles in the alternate segment)

= 30°

Now arc BP suntends ∠BOP at the centre and ∠BAP at the remaining part of the circle.

∠BOP = 2∠BAP

= 2×30°

= 60°

25. In the adjoining figure, ABCD is a cyclic quadrilateral.

The line PQ is the tangent to the circle at A. If CAQ : CAP = 1 : 2, AB bisects CAQ and AD bisects CAP, then find the measure of the angles of the cyclic quadrilateral. Also prove that BD is a diameter of the circle.

ABCD is a cyclic quadrilateral.

PAQ is the tangent to the circle at A.

∠CAD : ∠CAP = 1 : 2

AB and AD are the bisectors of ∠CAQ and ∠CAP respectively.

To prove:

(i) BD is the diameter of the circle.

(ii) Find the measure of the angles of the cyclic quadrilateral.

Proof:

(i) ∵ AB and AD are the bisectors of ∠CAQ and CAP respectively and

∠CAQ + ∠CAB = 180° (linear pair)

∴ ∠CAB = ∠CAD = 90°

⇒ ∠BAD = 90°

Hence, BD is at diameter of the circle.

(ii) ∴ ∠CAQ : ∠CAP = 1 : 2

Let ∠CAQ = x and ∠CAP = 2x

∴ x + 2x = 180°

⇒ 3x = 180°

⇒ x = 180°/3 = 60°

∴ ∠CAQ = 60° and ∠CAP = 120°

∵  PAQ is the tangent and AC is the chord of the circle.

∴ ∠ADC = ∠CAQ = 60°

Similarly,

∠ABC = ∠CAP = 120°

Hence ∠A = 90° , ∠B = 120°

∠C = 90° and ∠D = 60°

26. In a triangle ABC, the incircle (centre O) touches BC, CA and AB at P, Q and R respectively. Calculate (i) QOR (ii) QPR given that A = 60°.

OQ and OR are the radii and AC and AB are tangents.

OQ ⊥ AC and OR ⊥ AB

Now in the quad. AROQ

∠A = 60°, ∠ORA = 90° and ∠OQA = 90°

∴ ∠ROQ = 360° - (∠A + ∠ORA + ∠OQA)

= 360° - (60° + 90° + 90°)

= 360° - 240° = 120°

Now, arc RQ subtends ∠ROQ at the centre and ∠RPQ at the remaining part of the circle.

∴ ∠ROQ = 2∠RPQ

⇒ ∠RPQ = 1/2∠ROQ

⇒ ∠RPQ = ½ ∠ROQ

∠RPQ = ½ × 120°

= 60°

27. (a) In the figure given below, AB is a diameter. The tangent at C meets AB produced at Q, CAB = 34°. Find

(i) CBA

(ii) CQA

(b) In the figure (ii) Given below, AP ad BP are tangents o the circle with centre O. Given APB = 60°, calculate.

(i) AOB

(ii) OAB

(iii) ACB.

(a) AB is the diameter.

∴ ∠ACB = 90° and ∠CAB = 34°

∴ In Î”ABC,

∠ACB + ∠CAB + ∠CBA = 180°

⇒ 90° + 34° + ∠CBA = 180°

(b) (i) AP and BP are the tangents to the circle and OA and OB are radii on it.

∴ OA ⊥ AP and OB ⊥ BP

∴ ∠AOB = 180° - 60° = 120°

(ii) Now in Î”OAB,

OA = OB (radii of the same circle)

∴ ∠OAB = ∠OBA

= ½ (180° - 120°)

= ½ × 60°

= 30°

(iii) Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.

∴ ∠AOB = 2∠ACB

⇒ ∠ACB = ½ ∠AOB

⇒ ∠ACB = ½ × 120°

= 60°  [From (i)]

28. (a) In the figure (i) given below, O is the centre of the circumcircle of triangle XYZ. Tangents at X and Y intersect at T. Given XTY = 80° and XOZ = 140°, calculate the value of ZXY.

(b) In the figure (ii) given below, O is the centre of the circle and PT is the tangent to the circle at P. Given QPT = 30°, calculate (i) PRQ (ii) POQ.

(a) Join OY, OX and OY are the radii of the circle and XT and YT are the tangents to the circle.

∴ OX ⊥ XT and OY ⊥ YT

and ∠XYT = 80°

∴ ∠XOY = 180° - ∠XTY

= 180° - 80°

Now ∠ZOY = 360° - (∠XOZ + ∠XOY)

= 360° - (140° + 100°)

= 360° - 240°

= 120°

But arc ZY subtends ∠ZOY at the centre and ∠ZXY at the remaining part of the circle.

∴ ∠ZXY = 1/2∠ZOY

= ½ × 120°

= 60°

(b) (i) PT is tangent and OP is radius, then OP ⊥ PT.

∴ ∠OPT = 90°

But ∠QPT = 30°

∴ ∠OPQ = 90° - 30°

= 60°

∵ OP = OQ (radii of the same circle)

∴ ∠OQP = ∠OPQ = 60°

Hence in Î”OPQ,

∠POQ = 60°

(ii) Tale a point A on the circumference and join AP and AQ.

Now arc PQ subtends ∠POQ at the centre and ∠PAQ on the circumference of the circle.

∴ ∠PAQ = 1/2∠POQ

= ½ × 60°

= 30°

Now APRQ is a cyclic quadrilateral.

∴ ∠PAQ + ∠PRQ = 180°

⇒ 30° + PRQ = 180°

⇒ 30° + ∠PRQ = 180°

⇒ ∠PRQ = 180° - 30°

= 150°

29. Two chords AB, CD of a circle intersect internally at a point P. If

(i) AP = 6 cm, PB = 4 cm and PD = 3 cm, find PC.

(ii) AB = 12 cm, AP = 2 cm, PC = 5 cm, find PD.

(iii) AP = 5 cm, PB = 6 cm and CD = 13 cm, find CP.

In a circle, two chords AB and CD intersect each other at P internally.

∴  AP.PB = CP.PD

(i) When AP = 6 cm, PB = 4 cm, PD = 3 cm,

Then

⇒ 6 × 4 = CP × 3

⇒ CP = (6 × 4)/3 = 8 cm

Hence PC = 8 cm

(ii) When AB = 12 cm, AP = 2 cm, PC = 5 cm

PB = AB – AP

= 12 – 2

= 10 cm

∵ AP × PB = CP × PD

⇒ 2 × 10 = 5 × PD

⇒ PD = (2 × 10)/5 = 4 cm

(iii) When AP = 5 cm, PB = 6 cm, CD = 13 cm

Let CP = x, then PD = CD – CP

Or PD = 13 – x

∴  AP × PB = CP × PD

⇒ 5 × 6 = x (13 – x)

⇒ 30 = 13x – x2

⇒ x2 – 13x + 30 = 0

⇒ x2 – 10x – 3x + 30 = 0

⇒ x(x – 10) – 3(x – 10) = 0

⇒ (x – 10)(x – 3) = 0

Either x – 10 = 0, then x = 10

Or x – 3 = 0, then x = 3

∴ CP = 10 cm or 3 cm

30. (a) In the figure (i) given below, PT is a tangent to the circle. Find TP if AT = 16 cm and AB = 12 cm.

(b) In the figure given below, diameter AB and Chord CD of a circle meet at P. PT is a tangent to circle at T. CD = 7.8 cm, PD = 5 cm, PB = 4 cm. Find (i) AB . (ii) The length of tangent PT.

(a) PT is the tangent to the circle and AT is a secant.

PT2 = TA × TB

Now TA = 16 cm, AB = 12 cm

TB = AT – AB

= 16 – 12

= 4 cm

∴ PT2 = 16 + 4

= 64

= (8)2

⇒ PT = 8 cm or TP = 8 cm

(b) PT is tangent and PDC is secant out to the circle

∴ PT2 = PC×PD

⇒ PT2 = (5 + 7.8) × 5 = 12.8×5

⇒ PT2 = 64

⇒ PT = 8 cm

In Î”OTP,

PT2 + OT2 = OP2

⇒ 82 + x2 = (x + 4)2

⇒ 64 + x2 = x2 + 16 + 8x

⇒ 64 – 16 = 8x

⇒ 48 = 8x

⇒ x = 48/8 = 6 cm

∴ Radius = 6 cm

AB = 2 × 6 = 12 cm

31. PAB is secant and PT is tangent to a circle

(i) PT = 8 cm and PA = 5 cm, find the length o AB.

(ii) PA = 4.5 cm and AB = 13.5 cm, find the length of PT.

∴ PT is the tangent and PAB is the secant of the circle.

∴ PT2 = PA.PB

(i) PT = 8 cm, PA = 5 cm

∴ (8)2 = 5 × PB

⇒ 64 = 5 PB

⇒ PB = 64/5 = 12.8 cm

∴ AB = PB – PA

= 12.8 – 5.0

= 7.8 cm

(ii) PT2 = PA × PB

But PA = 4.5 cm, AB = 13.5 cm

∴ PB = PA + AB

= 4.5 + 13.5

= 18 cm

Now PT2 = 4.5 × 18

= 81

= (9)2

∴ PT = 9 cm.

32. In the adjoining figure, CBA is a secant and CD is tangent to the circle. If AB = 7 cm and BC = 9 cm, then

(i) Prove that Î”ACD ~ Î”DCB.

(ii) Find the length of CD.

In Î”ACD and Î”DCB

∠C = ∠C (common)

∠CAD = ∠CDB [Angle between chord and tangent is equal to angle made by chord in alternate segment.]

∴ Î”ACD ~ Î”DCB

∴ AC/DC = DC/BC

⇒ DC2 = AC × BC

= 16 × 9

= 144

⇒ DC = 12 cm

33. (a) In the figure (i) given below, PAB is secant and PT is tangent to a circle. If PA : AB = 1 : 3 and PT = 6 cm, find the length of PB.

(b) In the figure (ii) given below, ABC is an isosceles triangle in which AB = AC and Q is mid-point of AC. If APB is a secant, and AC is tangent to the circle at Q, prove that AB = 4 AP.

(a) In the figure (i),

PAB is secant and PT is the tangent to the circle.

PT2 = PA × PB

But PT = 6 cm, and PA : AB = 1 : 3

Let PA = x, then AB = 3x

∴ PB = PA + AB

= x + 3x

= 4x

Now PT2 = PA × PB

⇒ (6)2 = x(4x)

⇒ 36 = 4x2

⇒ x2 = 36/4 = 9

= (3)2

⇒ x = 3

∴ PB = 4x

= 4 × 3

= 12 cm

(b) In the figure (ii)

Î”ABC is an isosceles triangles in which AB = AC

Q is mid-point of AC.

APB is the secant and AC is the tangent to the circle at Q.

To prove: AB = 4 AP

Proof :

∵ AQ is the tangent and APB is the secant.

∴ AQ2 = AP × AB

(1/2 AC)2 = AP × AB

⇒ (1/2 AB)2 = AP × AB (∵AB = AC)

⇒ AB2/4 = AP × AB

⇒ AB2 = 4AP × AB

⇒ AB2/AB = 4 AP

⇒ AB = 4 AP

Hence, proved.

34. Two chords AB, CD of a circle intersect externally at a point P. If PA = PC, prove that AB = CD.

Given: Two chords AB and CD intersect each other at P outside the circle. PA = PC.

To prove: AB = CD

Proof: Chords AB and CD intersect each other at P outside the circle.

∴ PA×PB = PC×PD

But PA = PC (given) …(i)

∴ PB = PD …(ii)

Substracting (ii) from (i)

PA – PB = PC – PD

⇒ AB = CD Q.E.D.

35. (a) In the figure (i) given below, AT is tangent to a circle at A. If BAT = 45° and BAC = 65°, find ABC.

(b) In the figure (ii) given below, A, B and C are three points on a circle. The tangent at C meets BA produced at T. Given that ATC = 36° and ACT = 48°, calculate the angle subtended by AB at the centre of the circle.

(a) AT is the tangent to the circle at A and AB is the chord of the circle.

∴ ∠ACB = ∠BAT (Angles in the alternate segment)

= 45°

Now in Î”ABC,

∠ABC + ∠BAC + ∠ACB = 180° (Angles of a triangle)

⇒ ∠ABC + 65° + 45° = 180°

⇒ ∠ABC + 110° = 180°

∴ ∠ABC = 180° - 110°

= 70°

(b) Join OA, OB and CB. In Î”ATC,

Ext.∠CAB = ∠ATC + ∠TCA

= 36° + 48°

= 84°

But ∠TCA = ∠ABC (Angles in the alternate segment)

∴ ∠ABC = 48°

But in Î”ABC,

∠ABC + ∠BAC + ∠ACB = 180°

⇒ 48° + 84° + ∠ACB = 180°

⇒ 132° + ∠ACB = 180°

⇒ ∠ACB = 180° - 132°

⇒ ∠ACB = 48°

∵ Arc. AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.

∴ ∠AOB = 2∠ACB

= 2 × 48°

= 96°

36. In the adjoining figure Î”ABC is isosceles with AB = AC. Prove that the tangent at A to the circumcircle of Î”ABC is parallel to BC.

Given : Î”ABC is an isosceles triangle with AB = AC.

AT is the tangent to the circumcircle at A.

To prove: AT

To prove: AT || BC.

Proof:

In Î”ABC,

AB = AC (given)

∴ ∠C = ∠B (Angles opposite to equal sides)

But AT is the tangent and AC is the chord.

∴ ∠TAC = ∠B (Angles in the alternate segment)

But ∠B = ∠C (proved)

∴ ∠TAC = ∠C

But these are alternate angles

∴ AT || BC Q.E.D.

37. If the sides of a rectangle touch a circle, prove that the rectangle is a square.

Given : A circle touches the sides AB, CD and DA of a rectangle ABCD at P, Q, R and S respectively.

To prove: ABCD is a square.

Proof:

Tangents from a point to the circle are equal.

∴ AP = AS

Similarly BP = BQ

CR = CQ and DR = DS

AP + BP + CR + DR = AS + BQ + CQ + DS

⇒ AP + BP + CR + DR = AS + DS + BQ + CQ

⇒ AB + CD = AD + BC

But AB = CD and AD = BC (Opposite sides of a rectangle)

∴ AB + AB = BC + BC

⇒ 2 AB = 2 BC

⇒ AB = BC

∴ AB = BC = CD = DA

Hence, ABCD is a square.

38. (a) In the figure (i) given below, two circles intersect at A, B. From a point P on one of these circle, two line segments PAC and PBD are drawn, intersecting the other circle at C and D respectively. Prove that CD is parallel to the tangent at P.

(b) In the figure (ii) given below, two circles with centres C, C’ intersect at a, B and the point C lies on the circle with centre C’. PQ is a tangent to the circle with centre C’ at A. Prove that AC bisects PAB.

(a) Given: Two circles intersect each other at A and B.

From a point P on one circle, PAC and PBD are drawn.

From P, PT is a tangent drawn. CD is joined.

To prove : PT || CD.

Construction: Join AB.

Proof:

PT is tangent and PA is chord.

∴ ∠APT = ∠ABP (Angles in the alternate segments) …(i)

But BDCA is a cyclic quadrilateral

∴ Ext. ∠ABP = ∠ACD …(ii)

From (i) and (ii)

∠APT = ∠ACD

But these are alternate angles

∴ CD || PT

(b) Given : Two circles with centres C and C’ intersect each other at A and B. PQ is a tangent to the circle with centre C’ at A.

To Prove: AC bisects ∠PAB.

Construction: Join AB and CP.

Proof:

In Î”ACB,

AC = BC (radii of the same circle)

∴ ∠BAC = ∠ABC …(i)

PAQ is tangent and AC is the chord of the circle.

∠PAC = ∠ABC …(ii)

From (i) and (ii)

∠BAC = ∠PAC

Hence, AC is the bisector of ∠PAB

39. (a) In the figure (i) given below, AB is a chord of the circle with centre O, BT is tangent to the circle. If OAB = 32°, find the values of x and y.

(b) In the figure (ii) given below, O and O’ are centres of two circles touching each other externally at the point P. The common tangent at P meets a direct common tangent AB at M. Prove that:

(i) M bisect AB (ii) APB = 90°

(a) AB is a chord of a circle with centre O.

BT is a tangent to the circle and ∠OAB = 32°

∴ In Î”OAB,

OA = OB  (radii of the same circle)

∴ ∠OAB = ∠OBA

∴ x = 32°

And ∠AOB = 180°- (x + 32°)

= 180° - 64°

= 116°

Now, arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.

∠AOB = 2∠ACB

⇒ ∠ACB = 1/2∠AOB

= ½ × 116°

= 58°

Thus, x= 32° and y = 58°

(b) Given: Two circles with centre O and O’ touch each other at P externally. From P, a common tangent is drawn and meets the common direct tangent AB at M.

To prove : (i) M bisects AB i.e., AM = MB.

(ii) ∠APB = 90°

Proof :

(i) From M, MA and MP are the tangents.

∴ MA = MP …(i)

Similarly,

MB = MP …(ii)

From (i) and (ii)

MA = MB

Or, M is the mid-point of AB.

(ii) ∠MAP + ∠MPB = ∠MPA + MPB

⇒ ∠APB = ∠MAP + MBP

But ∠APB + ∠MAP + ∠MBP = 180°(Angles of a triangle)

⇒ ∠APB + ∠APB = 180°

⇒ 2∠APB = 180°

⇒ ∠APB = 180°/2

= 90°

### Multiple Choice Questions

1. In the adjoining figure, O is the centre of the circle. If ABC = 20°, then AOC is equal to

(a) 20°

(b) 40°

(c) 60°

(d) 10°

(b) 40°

In the given figure,

Arc AC subtends ∠AOC at the centre and ∠ABC at the remaining part of the circle

∠AOC = 2∠ABC

= 2 × 20°

= 40°

2. In the adjoining figure, AB is a diameter of the circle. If AC = BC, then CAB is equal to

(a) 30°

(b) 60°

(c) 90°

(d) 45°

(d) 45°

In the given figure,

AB is the diameter of the circle and AC = BC

∠ACB = 90° (angle in a semi-circle)

AC = BC

∴ ∠A = ∠B

But ∠A + ∠B = 90°

∴ ∠A = ∠B = 90°/2 = 45°

∴ ∠CAB = 45°

3. In the adjoining figure, if DAB = 60°and ABD = 50° then ACB is equal to

(a) 60°

(b) 50°

(c) 70°

(d) 80°

(c) 70°

In the given figure,

∠DAB = 60°, ∠ABD = 50°

In Î”ADB, Î”ADB = 180° - (60° + 50°)

= 180° - 110°

= 70°

∠ACB = ∠ADB (angles in same segment)

= 70°

4. In the adjoining figure, O is the centre of the circle. If OAB = 40°, then ACB is equal to

(a) 50°

(b) 40°

(c) 60°

(d) 70°

(a) 50°

In the given figure, O is the centre of the circle.

In Î”OAB,

∠OAB = 40°

But ∠OBA = ∠OAB = 40°

(∵ OA = OB radii of the same circle)

∴ ∠AOB = 180° - (40° + 40°)

= 180° - 80°

= 100°

But arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle

∴ ∠ACB = ½ ∠AOB

= ½ × 100°

= 50°

5. ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ADC = 140°, then BAC is equal to

(a) 80°

(b) 50°

(c) 40°

(d) 30°

(b) 50°

ABCD is a cyclic quadrilateral,

AB is the diameter of the circle circumscribing it

∠ADC = 140°, ∠BAC = Join AC

∴ ∠ADC + ∠ABC = 180° (Opposite angles of the cyclic quadrilateral)

140° + ∠ABC = 180°

∠ABC = 180° - 140° = 40°

Now in Î”ABC,

∠ACB = 90° (angle in a semi-circle)

∴ ∠BAC = 90°- ∠ABC

= 90° - 40°

= 50°

6. In the adjoining figure, O is the centre of the circle. If BAO = 60°, then ADC is equal to

(a) 30°

(b) 45°

(c) 60°

(d) 120°

(c) 60°

In the given figure, O is the centre of the circle ∠BAO = 60°

In Î”ABO, OA = OB (radii of the same circle)

∴ ∠ABO = ∠BAO = 60°

Ext. ∠AOC = ∠BAO + ∠ABO

= 60° + 60°

= 120°

Now arc AC subtends ∠AOC at the centre ∠ADC at the remaining part of the circle

∴ ∠AOC = 2∠ADC

⇒ 2∠ADC = 120°

⇒ ∠ADC = 120°/2

= 60°

7. In the adjoining figure, O is the centre of the circle. If ∠AOB = 90° and ∠ABC = 30° , then CAO is equal to

(a) 30°

(b) 45°

(c) 90°

(d) 60°

(d) 60°

In Î”AOB,

∠AOB = 90°, ∠ABC = 30°

In Î”AOB, ∠AOB = 90°

OA = OB (radii of the same circle)

∴ ∠OAB = ∠OBA = 90° /2

= 45°

Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle

∴ ∠ACB = 1/2 ∠AOB

= ½ × 90°

= 45°

Now in Î”ACB, ∠ABC = 30°, ∠ACB = 45°

∠BAC = 180°- (30°+ 45°)

= 180° - 75°

= 105°

But ∠OAB = 45°

∠CAO = 105°  - 45°

= 60°

8. In the adjoining figure, O is the centre of a circle. If the length of chord PQ is equal to the radius of the circle, then PRQ is

(a) 60°

(b) 45°

(c) 30°

(d) 15°

(c) 30°

In the given figure, O is the centre of the circle

Chord PQ = radius of the circle

Î”OPQ is an equilateral triangle

∴ ∠POQ = 60°

Arc PQ subtends ∠POQ at the centre and

∴ ∠PRQ at the remaining part of the circle

∴ ∠PRQ = 1/2∠POQ

= ½ × 60°

= 30°

9. In the adjoining figure, if O is the centre of the circle then the value of x is

(a) 18°

(b) 20°

(c) 24°

(d) 36°

(a) 18°

In the given figure, O is the centre of the circle.

Join OA.

∠ADB = ∠ACB = 2x (Angles in the same segment)

Are AB subtends ∠AOB at the centre and ∠ADB at the remaining part of the circle

∠AOB = 2∠AOB

= 2 × 2x

= 4x

In Î”OAB,

∠OAB = ∠OBA = 3x (OA = OB)

Sum of angles of Î”OAB = 180°

⇒ 3x + 3x + 4x = 180°

⇒ 10x = 180°

⇒ x = 180°/10

= 18°

∴ x = 18°

10. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is

(a) 7 cm

(b) 12 cm

(c) 15 cm

(d) 24.5 cm

(a) 7 cm

From Q, length of tangent PQ to the circle = 24 cm

And QO = 25 cm

∵ PQ is tangent and OP is radius

∴ OP ⊥ PQ

Now in right Î”OPQ

OQ2 = OP2 + PQ2

(25)2 = OP2 + (24)2

⇒ OP2 = 252 – 242 = 625 – 576

⇒ OP2 = 49 = (7)2

⇒ OP = 7 cm

∴ Radius of the circle = 7 cm

11. From a point which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is

(a) 60 cm2

(b) 65 cm2

(c) 30 cm2

(d) 32.5 cm2

(a) 60 cm2

Let point P is 13 cm from O, the centre of the circle

Radius of the circle (OQ) = 5 cm

PQ and PR are tangents from P to the circle

Join OQ and OR

∵ PQ is tangent and OQ is the radius

∴ PQ2 = OP2 – OQ2

= 132 – 52

= 169 – 25

= 144

= (12)2

∴ PQ = 12 cm

Now area of Î”OPQ = ½ PQ × OQ

= (1/2 × base × height)

= ½ × 12 × 5

= 30 cm2

∴ area of quad. PQOR = 2 × 30

= 60 cm2

12. If angle between two radii of a circle is 130°, the angle between the tangents at the ends of the radii is

(a) 90°

(b) 50°

(c) 70°

(d) 40°

(b) 50°

Angles between two radii OA and OB = 130°

From A and B, tangents are drawn which meet at P

∵ OA radius and AP is tangent to the circle

∴ ∠OAP = 90°

Similarly ∠OBP = 90°

∴ ∠AOB + ∠APB = 180°

⇒ 130° + ∠APB = 180°

∠APB = 180°- 130°

= 50°

13. In the adjoining figure, PQ and PR are tangents from P to a circle with centre O. If POR = 55°, then QPR is

(a) 35°

(b) 55°

(c) 70°

(d) 80°

(c) 70°

In the given figure,

PQ and PR are the tangents to the circle from appoint P outside it

∠POR = 55°

∵ OR is radius and PR is tangent

∴ OR ⊥ PR

∴ In Î”OPR

∠OPR = 90° - 55° = 35°

∠OPR = 2 × 35°

= 70°

14. If tangents PA and PB from an exterior point P to a circle with centre O are inclined to each other at an angle of 80°, then POA is equal to

(a) 50°

(b) 60°

(c) 70°

(d) 100°

(a) 50°

Length of tangents PA and PB to the circle from a point P outside the circle with centre O, and inclined an angle of 80°

∵ OA is radius and AP is tangent

∴ ∠OAP = 90° and ∠OPA = ½ ∠APB

= ½ × 80°

= 40°

∴ ∠POA = 90° - 40°

= 50°

15. In the adjoining figure, PA and PB are tangents from point P to a circle with centre O. If the radius of the circle is 5 cm and PA PB, then the length OP is equal to

(a) 5 cm

(b) 10 cm

(c) 7.5 cm

(d) 5√2 cm

(d) 5√2 cm

In the given figure,

PA and PB are tangents to the circle with centre O.

Radius of the circle is 5 cm, PA ⊥ PB.

OA is radius and PA is tangent to the circle

∴ OA ⊥ PA

∵ ∠APB = 90° (∵ PA ⊥ PB)

∴ ∠APO = 90° × ½ = 45°

∴ ∠APO = 90° - 45°

= 45°

i.e., OA = AP = 5 cm

16. At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A is

(a) 4 cm

(b) 5 cm

(c) 6 cm

(d) 8 cm

(d) 8 cm

AB is the diameter of a circle with radius 5 cm

At A, XAY is a tangent to the circle

CD || XAY at a distance of 8 cm from A

Join OC

In right Î”OEC,

OE = 8 – 5 = 3 cm

OC = 5 cm

= 4 cm

∴ CD = 2 × CE

= 2 × 4

= 8 cm

17. If radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is tangent to the other is

(a) 3 cm

(b) 6 cm

(c) 9 cm

(d) 1 cm

(b) 6 cm

Radii of two concentric circles are 4 cm and 5 cm

AB is a chord of the bigger circle which is tangent to the smaller circle at C.

Join OA, OC

OC = 4 cm, OA = 5 cm

And OC ⊥ ACB

∴ In right Î”OAC

OA2 = OC2 + AC2

⇒ 52 = 42 + AC2

⇒ 25 = 16 + AC2

⇒ AC2 = 25 – 16

= 9

= (3)2

∴ AC = 3 cm

∴ Length of chord AB = 2 × AC

= 2 × 3

= 6 cm

18. In the adjoining figure, AB is a chord of the circle such that ACB = 50°. If AT is tangent to the circle at the point A, then BAT is equal to

(a) 65°

(b) 60°

(c) 50°

(d) 40°

(c) 50°

In the given figure, AB is a chord of the circle such that ∠ACB = 50°

AT is tangent to the circle at A

AT is tangent and AB is a chord

∠ACB = ∠BAT = 50° (Angles in the alternate segments)

19. In the adjoining figure, O is the centre of a circle and PQ is a chord. If the tangent PR at P makes an angle of 50° with PQ, then POQ is

(a) 100°

(b) 80°

(c) 90°

(d) 75°

(a) 100°

In the given figure, O is the centre of the circle.

PR is tangent and PQ is chord ∠RPQ = 50°

OP is radius and PR is tangent to the circle

∠RPQ = 50°

∵ OP is radius and PR is tangent to the circle

∴ OP ⊥ PR

But ∠OPQ + RPQ = 90°

⇒ ∠OPQ + 50° = 90°

⇒ ∠OPQ = 90° - 50°

= 40°

∵ OP = OQ (radii of the same circle)

∴ ∠OQP = ∠OPQ = 40°

And ∠POQ = 180 ̊ - (∠OPQ + ∠OQP)

= 180° - (40°+40°)

= 180° - 80°

= 100°

20. In the adjoining figure, PA and PB are tangents to a circle with centre O. If APB = 50°,  then OAB is equal to

(a) 25°

(b) 30°

(c) 40°

(d) 50°

(a) 25°

In the given figure,

PA and PB are tangents to the circle with centre O.

∠APB = 50°

But ∠AOB + ∠APB = 180°

∠AOB + 50° = 180°

∠AOB = 180° - 50°

= 130°

In Î”OAB,

OA = OB (radii of the same circle)

∠OAB = ∠OBA

But ∠OAB + ∠OBA = 180° - ∠AOB

= 180° - 130°

= 50°

∠OAB = 50 ̊/2

= 25°

21. In the adjoining figure, sides BC, CA and AB of Î”ABC touch a circle at point D, E and F respectively. If BD = 4 cm, DC = 3 cm and CD = 8 cm, then the length offside AB is

(a) 12 cm

(b) 11 cm

(c) 10 cm

(d) 9 cm

(d) 9 cm

In the given figure, sides BC, CA and AB of Î”ABC touch a circle at D, E and F respectively.

BD = 4 cm, DC = 3 cm and CA = 8 cm

∵ BD and BF are tangents to the circle

∴ BF = BD = 4 cm

Similarly, CD = CE = 3 cm

∴ AE = AC – CE = 8 – 3 = 5 cm

And AF = AE = 5 cm

Now AB = AF + BF

= 5 + 4

= 9 cm

22. In the adjoining figure, sides BC, CA and AB of Î”ABC touch a circle at the points P, Q and R respectively If PC = 5 cm, AR = 4 cm and RB = 6 cm, then the perimeter of Î”ABC is

(a) 60 cm

(b) 45 cm

(c) 30 cm

(d) 15 cm

(c) 30 cm

In the given figure, sides BC, CA and AB of Î”ABC touch a circle at P, Q and R respectively

PC = 5 cm, AR = 4 cm, RB = 6 cm

∴ AR and AQ are tangents to the circles

∴ AQ = AR = 4 cm

Similarly CQ = CP = 5 cm

And BP = BR = 6 cm

Now AB = AR + BR

= 4 + 6

= 10 cm

BC = BP + CP

= 6 + 5

= 11 cm

AC = AQ + CQ

= 4 + 5

= 9 cm

∴ Perimeter of the Î”ABC = AB + BC + CA

= 10 + 11 + 9

= 30 cm

23. PQ is a tangent to a circle at point P. Centre of circle is O. If Î”OPQ is an isosceles triangle, then QOP is equal to

(a) 30°

(b) 60°

(c) 45°

(d) 90°

(c) 45°

PQ is tangent to the circle at point P centre of the circle is O.

Î”OPQ is an isosceles triangle

OP = PQ

∵ OP is radius and PQ is tangent to the circle

∴ OP ⊥ PQ i.e., ∠OPQ = 90°

∵ OP = PQ (∵ Î”OPQ is an isosceles triangle)

∴ ∠QOP = ∠PQO = 90°/2

= 45°

24. In the adjoining figure, PT is a tangent at T to the circle with centre O. If TPO = 25°, then the value of x is

(a) 25°

(b) 65°

(c) 115°

(d) 90°

(c) 115°

In the given figure, PT is the tangent at T to the circle with centre O.

∠TPO = 25°

OT is the radius and TP is the tangent

∴ OT ⊥ TP

∴ In Î”OPT

∠TOP + OPT = 90°

⇒ ∠TOP + 25° = 90°

⇒ ∠TOP = 90° - 25°

= 65°

∠TOP + x = 180° (Linear pair)

65° + x = 180°

⇒ x = 180° - 65°

= 115°

∴ x = 115°

25. In the adjoining figure, PA and PB are tangents at points A and B respectively to a circle with centre O. If C is a point on the circle and APB = 40°, then ACB is equal to

(a) 80°

(b) 70°

(c) 90°

(d) 140°

(b) 70°

In the given figure,

PA and PB are tangents to the circle at A and B respectively

C is a point on the circle and ∠APB = 40°

But ∠APB + ∠AOB = 180°

⇒ 40° + ∠AOB = 180°

⇒ ∠AOB = 180° - 40° = 140°

Now arc AB subtends ∠AOB at the centre and ∠ACB is on the remaining part of the circle

∴ ∠ACB = ½ ∠AOB

= ½ × 140°

= 70°

26. In the adjoining figure, two circles touch each other at A. BC and AP are common tangents to these circles. If BP = 3.8 cm, then the length of BC is equal to

(a) 7.6 cm

(b) 1.9 cm

(c) 11.4 cm

(d) 5.7 cm

(a) 7.6 cm

In the given figure, two circles touch each other at A.

BC and AP are common tangents to these circles

BP = 3.8 cm

∵ PB and PA are the tangents to the first circle

∴ PB = PA = 3.8 cm

Similarly PC and PA are tangents to the second circle

∴ PA = PC = 3.8 cm

BC = PB + PC

= 3.8 + 3.8

= 7.6 cm

27. In the adjoining figure, if sides PQ, QR, RS and SP of a quadrilateral PQRS touch a circle at points A, B, C and D respectively, then PD + BQ is equal to

(a) PQ

(b) QR

(c) PS

(d) SR

(a) PQ

In the given figure, sides PQ, QR, RS and SP of a quadrilateral PQRS touch a circle at the points A, B, C and D respectively

PD and PA are the tangents to the circle

∴ PA = PD …(i)

Similarly, QA and QB are the tangents

∴ QA = QB …(ii)

Now PD + BQ = PA + QA = PQ [From (i) and (ii)]

28. In the adjoining figure, PQR is a tangent at Q to a circle. If AB is a chord parallel to PR and BQR = 70°, then AQB is equal to

(a) 20°

(b) 40°

(c) 35°

(d) 45°

(b) 40°

In the given figure, PQR is a tangent at Q to a circle.

Chord AB || PR and ∠BQR = 70°

BQ is chord and PQR is a tangent

∠BQR = ∠A (Angles in the alternate segments)

∵ AB || PQR

∴ ∠BQR = ∠B (alternate angles)

∴ ∠A = ∠B = 70°

∴ ∠AQB + ∠A + ∠B = 180° (Angles of a triangle)

⇒ ∠AQB + 70° + 70° = 180°

⇒ ∠AQB + 140° = 180°

∴ ∠AQB = 180° - 140°

= 40°

29. Two chords AB and CD of a circle intersect externally at a point P. If PC = 15 cm, CD = 7 cm and AP = 12 cm, then AB is

(a) 2 cm

(b) 4 cm

(c) 6 cm

(d) none of these

(a) 2 cm

In the given figure,

Two chords AB and CD of a circle intersect externally at P.

PC = 15 cm, CD = 7 cm, AP = 12 cm

Join AC and BD

In Î”APC and Î”BPD

∠P = ∠P (common)

∠A = ∠BDP {Ext. of a cyclic quad. is equal to its interior opposite angles}

∴ Î”APC ~ Î”BPD (AA axiom)

PA/PD = PC/PB

PA.PB = PC.PD

12.PB = 15 × 8 (PD = 15 – 7 = 8)

PB = (15 × 8)/12 = 10

∴ AB = AP – PB

= 12 – 10

= 2 cm

### Chapter Test

1. (a) In the figure (i) given below, triangle ABC is equilateral. Find BDC and BEC.

(b) In the figure (ii) given below, AB is a diameter of a circle with center O. OD is perpendicular to AB and C is a point on the arc DB. Find BAD and ACD

Solution

(a) Triangle ABC is an equilateral triangle.

Each angle = 60o

∠A = 60o

But ∠A = ∠D (Angles in the same segment)

∠D = 600

Now, ABEC is a cyclic quadrilateral,

∠A = ∠E = 180o

⇒ 60+ ∠E = 180o

⇒ 60+ ∠E = 180o (∠E = 180– 60o)

⇒ ∠E = 120o

Hence, ∠BDC = 60and ∠BEC = 120o

(b) AB is diameter of circle with centre O. OD ⊥ AB and C is a point on arc DB.

(i) In ∆AOD, ∠AOD = 90°

OA = OD (radii of the semi–circle)

But ∠OAD + ∠ODA = 90o

⇒ ∠OAD + ∠ODA = 90o

⇒ 2∠OAD = 90o

⇒ ∠OAD = 90o/2 = 45°

Or ∠BAD = 45o

(ii) Arc AD subtends ∠AOD at the centre and ∠ACD at the remaining part of the circle.

∠AOD = 2 ∠ACD

⇒ 90o = 2 ∠ACD (OD ⊥ AB)

⇒ ∠ACD = 90o/2 = 45o

2. (a) In the figure given below, AB is a diameter of the circle. If AE = BE and ADC = 118°, find (i) BDC (ii) CAE

(b) In the figure given below, AB is the diameter of the semi-circle ABCDE with centre O. If AE = ED and BCD = 140°, find AED and EBD. Also Prove that OE is parallel to BD.

Solution:

(a) (i) Join DB, CA and CB. ∠ADC = 118° (given) and ∠ADB = 90° (Angles in a semi-circle)

= 118° – 90o = 28o (ABCD is a cyclic quadrilateral)

(ii) ∠ADC + ∠ABC = 180o

⇒ 118+ ∠ABC = 180o

⇒ ∠ABC = 180– 118o = 62o

But in ∆AEB,

∠AEB = 90(Angles in a semi-circle)

∠EAB = ∠ABE (AE = BE)

∠EAB + ∠ABE = 90o

⇒ ∠EAB = 90o × ½ = 45o

⇒ ∠CBE = ∠ABC + ∠ABE

= 62+ 45= 107o

But AEBD is a cyclic quadrilateral

∠CAE + ∠CBE = 180o

⇒ ∠CAE + 107= 180o

⇒ ∠CAE = 180o – 107= 73o

(b) AB is the diameter of semi-circle ABCDE

With center O.AE = ED and ∠BCD = 140o

In cyclic quadrilateral EBCD.

(i) ∠BCD + ∠BED = 180 ̊

⇒ 140 ̊ + ∠BED = 180 ̊

⇒ ∠BED = 180 ̊ – 140o = 400

But ∠AED = 90(Angles in a semi circle)

⇒ ∠AED = ∠AEB + ∠BED

= 90+ 40o = 130o

(ii) Now in cyclic quadrilateral AEDB

∠AED + ∠DBA = 180o

⇒ 130+ ∠DBA =180o

⇒ ∠BDA = 180o – 130o = 50o

Chord AE = ED (given)

∠DBE = ∠EBA

But ∠DBE + ∠EBA = 50o

⇒ DBE + ∠DBE = 50o

⇒ 2∠DBE = 50o

⇒ ∠DBE = 25or ∠EBD = 25o

In ∆OEB,

OE = OB (radii of the same circle)

∠OEB = ∠EBO = ∠DBE

But these are ultimate angles

OE ∥ BD

3. (a) In the figure (i) given below, O is the centre of the circle. Prove that AOC = 2 (ACB + BAC).

(b) In the figure (ii) given below, O is the centre of the circle. Prove that x + y = z

(a) Given: O is the center of the circle.

To Prove : ∠AOC = 2 (∠ACB + ∠BAC).

Proof:

In ∆ABC,

∠ACB + ∠BAC + ∠ABC = 180° (Angles of a triangle)

∠ABC = 180– (∠ACB + ∠BAC) …(i)

In the circle, arc AC subtends ∠AOC at

The center and ∠ABC at the remaining part of the circle.

Reflex ∠AOC = 2 ∠ABC …(ii)

Reflex AOC = 2 {180– (ACB + BAC)}

But ∠AOC = 360o – 2(∠ACB + ∠BAC)

But ∠AOC = 360– reflex ∠AOC

= 360 ̊ – (360– 2(∠ACB + ∠BAC)

= 360– 360+ 2 (∠ACB + ∠BAC)

= 2 (∠ACB + ∠BAC)

Hence ∠AOC = 2 (∠ACB + ∠BAC)

(b) Given : in the figure, O is the center of the circle

To Prove : x + y = z.

Proof : Arc BC subtends ∠AOB at the center and ∠BEC at the remaining part of the circle.

∠BOC = 2 ∠BEC

But ∠BEC = ∠BDC (Angles in the same segment)

∠BOC = ∠BEC + ∠BDC …(i)

In ∆ABD

Ext. ∠y = ∠EBD +∠EBC

∠BEC = y - ∠EBD ...(ii)

Similarly in ∆ABD

Ext. ∠BDC = x + ∠ABD

= x + ∠EBD …(iii)

Substituting the value of (ii) and (iii) in (i)

∠BOC = y – ∠EBD + x + ∠EBD = x + y

z = x + y

4. (a) In the figure (i) given below, AB is the diameter of a circle. If DC is parallel to AB and CAB = 25°, find:

(ii) DAC

(b) In the figure (ii) given below, the centre O of the smaller circle lies on the circumference of the bigger circle. If APB = 70° and BCD = 60°, find :

(i) AOB

(ii) ACB

(iii) ABD

(a) AB is diameter and DC || AB,

∠CAB = 25°, Join AD, BD

∠BAC = ∠BDC (Angles in the same segment)

But ∠ADB = 90° (Angles in a semi-circle)

∠ADC = ∠ADB + ∠BAC = 90° + 25° = 115°

∵ DC || AB (given)

∴ ∠CAB = ∠ACD (alternate angles)

∴ ∠ACD = 25°

Now in Î”ACD,

∠DAC + ∠ADC + ∠ACD = 180° (Angles of a triangle)

⇒ ∠DAC + 115° + 25° = 180°

⇒ ∠DAC + 140° = 180°

⇒ ∠DAC = 180° - 140° = 40°

(b) (i) Arc AB subtends ∠AOB at the centre and ∠APB at the remaining part of the circle.

∴ ∠AOB = 2∠APB

= 2 ×70°

= 140°

(ii) AOBC is a cyclic quadrilateral.

∴ ∠AOB + ∠ACB = 180°

⇒ 140° + ∠ACB = 180°

⇒ ∠ACB = 180° - 140°

= 40°

(iii) In cyclic quadrilateral ABDC

∠ABD + ∠ACD = 180°

⇒ ∠ABD + ∠ACB + ∠BCD = 180°

∠ABD + 40° + 60° = 180°

∠ABD = 180° - 100° = 80°

(iv) ∠ADB = ∠ACB (Angles in the same segment)

∴ ∠ADB = 40°

5. (a) In the figure (i) given below, ABCD is a cyclic quadrilateral. If AB = CD, Prove that AD = BC.

(b) In the figure (ii) given below, ABC is an isosceles triangle with AB = AC. If ABC = 50°, find BDC and BEC.

(a) Given : ABCD is a cyclic quadrilateral AB = CD.

To prove : AD = BC.

Construction : Join AD and BC.

Proof :

In Î”ABD and Î”CBD

AB = CD (given)

BD = BD (Common)

∠BAD = ∠BCD (Angles in the same segment)

∴ Î”ABC ≅ Î”CBD (SSA axiom of congruency)

∴ BC = AD

(b) Given: Î”ABC is an isosceles triangle and ∠ABC = 50°

In Î”ABC, an isosceles triangle

⇒ ∠ACB = ∠ABC (∵ opp. ∠s of an isosceles Î”s)

⇒ ∠ACB = 50°

Also in Î”ABC

⇒ ∠ABC + ∠ACB + ∠BAC = 180° (Sum of an isosceles triangle is 180°)

∴ 50° + 50° + ∠BAC = 180°

⇒ ∠BAC = 180°- 100°

⇒ ∠BAC = 80° (Angles in the same segment)

Now ABEC is a cyclic quadrilateral

∴ ∠A + ∠E = 180°

⇒ 80 ̊ + ∠E = 180°

⇒ ∠E = 180° - 80°

∴ ∠E = 100°

Hence ∠BDC = 80° and ∠BEC = 180°

6. A point P is 13 cm from the centre of a circle. The length of the tangent drawn from P to the circle is 12 cm. Find the distance of P from the nearest point of the circle.

Join OT, OP = 13 cm and TP = 12 cm

∵ OT is the radius

∴ OT ⊥ TP

Now in right Î”OTP

OP2 = OT2 + TP2

⇒ (13)2 = OT2 + (12)2

⇒ 169 = OT2 + 144

⇒ OT2 = 169 – 144

= 25

= (5)2

∴ OT = 5 cm

The nearest point A from P to cut circle over OA = radius of the circle = 5 cm.

∴ AP = OP – OA

= 13 – 5

= 8 cm

7. Two circles touch each other internally. Prove that the tangents drawn to the two circles from any point on the common tangent are equal in length.

Given: Two circles with centre O and O’ touch each other internally at P.

PT is any point on the common tangent of circles at P. From T, TA and TB are the tangents drawn to two circles.

To prove: TA = TB.

Proof : From T, TA and TP are the tangents to the first circle.

∴ TA = TP …(i)

Similarly, from T, TB and TP are the tangents to the second circle.

∴ TB = TP …(ii)

From (i) and (ii)

TA = TB

8. From a point outside a circle, with centre O, tangents PA and PB are drawn. Prove that

(i) AOP = BOP

(ii) OP is the perpendicular bisector of the chord AB.

Given: From a point P, outside the circle with centre O.

PA and PB are the tangents to the circle,

OA, OB and OP are joined.

To prove: (i) ∠AOP = ∠BOP

(ii) OP is the perpendicular bisector of the chord AB.

Construction: Join AB which intersects OP at M.

Proof:

In right Î”OAP and OBP

Hypotenuse OP = OP (common)

Side OA = OB (radii of the same circle)

∴ Î”OAP ≅ Î”OBP (R.H.S axiom of congruency)

∴ ∠AOP = ∠BOP (C.P.C.T.)

And ∠APO = ∠BPO (C.P.C.T)

Now in Î”APM and Î”BPM,

PM = PM (common)

∠APM = ∠BPM (proved)

AP = BP (tangents from P to the circle)

∴ Î”APM ≅ Î”BPM (SAS axiom of congruency)

∴ AM = BM (C.P.C.T)

And ∠AMP = ∠BMP

But ∠AMP + ∠BMP = 180°

∴ ∠AMP = ∠BMP = 90°

∴ OP is perpendicular bisector of AB at M.

9. (a) The figure given below shows two circles with centres A, B and a transverse common tangent to these circles meet the straight line AB in C.

Prove that:

AP : BQ = PC : CQ.

(b) In the figure (ii) given below, PQ is a tangent to the circle with centre O and AB is a diameter of the circle. If QA is parallel to PO, prove that PB is tangent to the circle.

(a) Given: Two circles with centres A and B and a transverse common tangent to these circles meets AB at C.

To prove : AP : BQ = PC : CQ

Proof : In Î”APC and Î”BQC

∠PCA = ∠QCB (vertically opposite angles)

∠APC = ∠BQC (each 90°)

∴ Î”APC ~ Î”BQC

∴ AP/BQ = PC/CQ

⇒ AP : BQ = PC : CQ

(b) Given : In the figure, O is the centre of the circle. AB is diameter. PQ is the tangent and QA || PO

To prove : PB is tangent to the circle

Construction: Join OQ

Proof : In Î”OAQ

OQ = OA (Radii of the same circle)

∴ ∠OQA = ∠POB (Corresponding angles)

And ∠OQA = ∠QOP (Alternate angles)

But ∠QAO = ∠OQA (Proved)

∠POB = ∠QOB

Now, in Î”OPQ and Î”OBP

OP = OP (Common)

OQ = OB (Radii of the same circle)

∠BOP = ∠POB (Proved)

∴ Î”OPQ ≅ Î”OBP (SAS axiom)

∴ ∠OQP = ∠OBP (c.p.c.t)

But ∠OQP = 90° (∵ PQ is tangent and OQ is the radius)

∴ ∠OBP = 90°

∴ PB is the tangent of the circle.

10. In the figure given below, two circles with centres A and B touch externally. PM is a tangent to the circle with centre A and QN is a tangent to the circle with centre B. If PM = 15 cm, QN = 12 cm, QN = 12 cm, PA = 17 cm and QB = 13 cm, then find the distance between the centres A and B of the circles.

In the given figure, two chords with centre A and B touch externally.

PM is a tangent to the circle with centre A

And QN is tangent to the circle with centre B.

PM = 15 cm, QN = 12 cm, PA = 17 cm, QB = 13 cm.

We have to find AB.

AM is radius and PM is tangent

∴ AM ⊥ PM

Similarly, BN ⊥ NQ

Now in right Î”APM,

AP2 = AM2 + PM2

⇒ 172 = AM2 + 152

⇒ AM2 = 172 – 152

= 289 – 225

= 64

= (8)2

∴  AM = 8 cm

Similarly in right Î”BNQ

BQ2 = BN2 + NQ2

⇒ 132 = BN2 + 122

⇒ 169 = BN2 + 144

BN2 = 169 – 144

= 25

= (5)2

∴ BN = 5 cm

Now AB = AM + BN (AR = AM and BR = BN)

= 8 + 5

= 13 cm

11. Two chords AB, CD of a circle intersect externally at a point P. If PB = 7 cm, AB = 9 cm and PD = 6 cm, find CD.

∵ AB and CD are two chords of a circle which intersect each other at P, outside the circle.

∴ PA × PB = PC × PD.

PB = 7 cm, AB = 9 cm, PD = 6 cm

AP = AB + BP

= 9 + 7

= 16 cm

∴ PA × PB = PC × PD

⇒ 16 × 7 = PC × 6

PC = (16 × 7)/6

= 56/3 cm

∴ CD = PC – PD

= 56/3 – 6

= 38/3

= 12.2/3 cm

12. (a) In the figure (i) given below, chord AB and diameter CD of a circle with centre O meet at P. PT is tangent to the circle at T. If AP = 16 cm, AB = 12 cm and DP = 2 cm, find the length of PT and the radius of the circle.

(b) In the figure (ii) given below, chord AB and diameter CD of a circle meet at P. If AB = 8 cm, BP = 6 cm and PD = 4 cm, find the radius of the circle. Also, find the length of the tangent drawn from P to the circle.

Given: (a) AB is a chord of a circle with centre O and PT is tangent and CD is the diameter of thecircle which meet at P.

AP = 16 cm, AB = 12 cm, OP = 2 cm

∴ PB = PA – AB = 16 – 12

= 4 cm

∵ ABP is a secant and PT is tangent.

∴ PT2 = PA × PB

= 16 × 4 = 64

= (8)2

⇒ PT = 8 cm

Again PT2 = PD × PC

⇒ (8)2 = 2 × PC

⇒ PC = (8 × 8)/2

⇒ PC = 32 cm

∴ CD = PC – PD

= 32 – 2

= 30 cm.

Radius of the circle = 30/2

= 15 cm

(b) Chord AB and diameter CD intersect each other at P outside the circle. AB = 8 cm, BP = 6 cm, PD = 4 cm.

PT is the tangent to the circle drawn from P.

∵ Two chords AB and CD intersect each other at P outside the circle.

PA = AB +PB

= 8 + 6

= 14 cm

∴ PA × PB = PC × PD

⇒ 14 × 6 = PC × 4

⇒ PC = (14 × 6)/4

= 84/4

= 21 cm

∴ CD = PC – PD

= 21 – 4

= 17 cm

∴ Radius of the circle = 17/2

= 8.5 cm

(ii) Now PT is the tangent and ABP is secant.

∴ PT2 = PA × PB

= 14 × 6

= 84

13. In the adjoining figure given below, chord AB and diameter PQ of a circle with centre O meet at X. If BX = 5 cm, OX = 10 cm and the radius of the circle is 6 cm, compute the length of AB. Also, find the length of tangent drawn from X to the circle.

Chord AB and diameter PQ meet at X on producing outside the circle.

BX = 5 cm, OX = 10 cm and radius of the circle = 6 cm

∴ XP = XO + OP = 10 + 6 = 16 cm

XQ = XO – OQ

= 10 – 6

= 4 cm

∴ XB.XA = XP.XQ

⇒ 5.XA = 16 × 4

⇒ XA = (16 × 4)/5

= 64/5 cm

XA = 12.4/5 cm

∴ AB = XA – XB

= 12.4/5 – 5

= 7.4/5 cm

Now ∵ XT is the tangent of the circle.

∴ XT2 = XP. XQ

= 16 × 4

= 64

= (8)2

XT = 8 cm

14. (a) In the figure (i) given below, CBP = 40°, CPB = q° and DAB = p°. Obtain an equation connecting p and q. If AC and BD meet at Q so that AQD = 2 q° and the points C, P, B and Q are concyclic, find the values of p and q.

(b) In the figure (ii) given below, AC is a diameter of the circle with centre O. If CD || BE, AOB = 130° and ACE = 20°, find :

(i) BEC

(ii) ACB

(iii) BCD

(iv) CED

(a) (i) Given: ABCD is a cyclic quadrilateral.

Ext. ∠PBC = ∠ADC

⇒ 40° = ∠ADC

p + q + ∠ADP = 180°

p + q = 180° - ∠ADP

= 180° - ∠ADC

= 180° - 40°

= 140°

∴ p + q = 140°

(ii) ∵ C, P, B, Q are concyclic

∴ ∠CPB + ∠CQB = 180°

⇒ q + 2q = 180° (∵ ∠CQB = ∠DQA)

⇒ 3q = 180°

∴ q = 180°/3 = 60°

But p + q = 140°

∴ p + 60° = 140°

⇒ p = 140° - 60°

= 80°

Hence p = 80°, q = 60°

(b) ∠AOB = 130 ̊

But ∠AOB + ∠BOC = 180° (Linear pair)

⇒ 130° + ∠BOC = 180°

∠BOC = 180° - 130°

= 50°

(i) Now arc BC subtends ∠BOC at the centre and ∠BEC at the remaining part of the circle.

∠BEC = 1/2∠BOC

= ½ × 50°

= 25°

(ii) Similarly arc AB subtends ∠AOB at the centre and ∠ACE at the remaining part of the circle

∠ACB = 1/2∠AOB

= ½ × 130°

= 65°

(iii) ∵ CD || EB

∴ ∠ECD = ∠CEB (alternate angles)

= 25°

∴ ∠BCD = ∠ACB + ∠ACE + ∠ECD

= 65° + 20° + 25°

= 110°

(iv) ∵ EBCD is a cyclic quadrilateral

∴ ∠CED + ∠BCD = 180°

⇒ ∠CED + ∠BEC + ∠BCD = 180°

⇒ ∠CED + 25° + 110° = 180°

⇒ ∠CED + 135° = 180°

∴ ∠CED = 180° - 135°

= 45°

15. (a) In the figure (i) given below, APC, AQB and BPD are straight lines.

(i) Prove that ADB + ACB = 180°

(ii) If a circle can be drawn through A, B, C and D, Prove that it has AB as a diameter

(b) In the figure (ii) given below, AQB is a straight line. Sides AC and BC of Î”ABC cut the circles at E and D respectively. Prove that the point C, E, P and D are concyclic.

(a) Given : In the figure, APC,

AQB and BPD are straight lines.

(i) To prove : (i) ∠ADB + ∠ACB = 180°

(ii) If a circle is drawn through A, B, C and D, then AB is a diameter.

Construction: Join PQ.

Proof :

In cyclic quad. AQPD

∠ADP + ∠AQB = 180°

But ∠AQB = ∠PCB (Ext. angle of a cyclic quad. is equal to its interior opposite angle)

∴ ∠ADP + ∠PCB = 180°

⇒ ∠ADB + ∠ACB = 180°

(ii) If A, B, C and D are concyclic then

∠ADB = ∠ACB (Angles in the same segment)

But ∠ADB + ∠ACB = 180° (Proved in (i))

∴ ∠ADB = ∠ACB = 90°

But these are angles on one side of AB.

∴ AB is the diameter of the circle.

(b) Given: AQB is a straight line. Sides AC and BC of Î”ABC cut the circles at E and D respectively.

To prove: C, E, P, D are concyclic.

Construction: Join PE, PD and PQ.

Proof :

In cyclic quad. AQPE,

∠A + ∠EPQ = 180°

⇒ ∠EPQ = 180° - ∠A ...(i)

Similarly PQBD is cyclic quadrilateral

∴ ∠QPD = 180° - ∠B

But ∠EPD + ∠EPQ + QPD = 360°(Angles at a point)

∠EPD + 180°- ∠A + 180° - ∠B = 360°

⇒ ∠EPD = ∠A + ∠B

Adding ∠C both sides,

∠EPD + ∠C = ∠A + ∠B + ∠C = 180°

∴ EPDC is a cyclic quadrilateral.

Hence E, P, D and C are concyclic.

16. (a) In the figure (i) given below, chords AB, BC and CD of a circle with centre o are equal. If BCD = 120°, find

(i) BDC

(ii) BEC

(iii) AEC

(iv) AOB.

Hence, Prove that AOAB is equilateral.

(b) In the figure (ii) given below, AB is a diameter of a circle with centre O. The chord BC of the circle is parallel to the radius OD and the lines OC and BD meet at E. Prove that

(i) CED = 3 CBD

(ii) CD = DA

(a) In Î”BCD, BC = CD

∠CBD = ∠CDB

But ∠BCD + ∠CBD + ∠CDB = 180° (∵ Angles of a triangle)

⇒ 120° + ∠CBD + ∠CBD = 180°

⇒ 120° + 2∠CBD = 180°

⇒ 2∠CBD = 180° - 120° = 60°

∴ ∠CBD = 30° and ∠CDB = 30°

∠BEC = ∠CDB (Angles in the same segment)

∴ ∠BEC = 30°

∵ CB = AB

∴ ∠BEC = ∠AEB (equal chords subtend equal angles)

∴ ∠AEB = 30°

Arc AB subtends ∠AOB at the centre and ∠AEB at the remaining part of the circle.

∴ ∠AOB = 2∠AEB

= 2 × 30°

= 60°

Now in Î”AOB

OA + OB (radii of the same circle)

∴ ∠OAB = ∠OBA

But ∠OAB + ∠OBA + ∠AOB = 180°

⇒ ∠OAB + ∠OAB + 60° = 180°

⇒ 2∠OAB = 180° - 60° = 120°

∴ ∠OAB = 60°

∠OAB = ∠OBA = ∠AOB = 60°

Hence OAB is an equailateral triangle.

(b) Given : In a circle with centre O, AB is diameter and chord BC || radius OD, OC and BD intersect each other at E.

To prove : (i) ∠CED = 3∠CBD

(ii) CD = DA.

Construction: CD, DA are joined.

Proof :

Arc CD subtends ∠COD at the centre and ∠CBD at the remaining part of the circle.

∴ ∠COD = 2∠CBD …(i)

∵ BC || OD

∴ ∠CBD = ∠BDO ...(ii)

In Î”DOE,

Ext. ∠BEO = ∠EDO + ∠EOD

= ∠BDO + ∠COD

= ∠CBD + 2∠CBD

From (i) and (ii)

= 3 ∠CBD

But ∠CED = ∠BEO (vertically opposite angles)

∴ ∠CED = 3∠CBD

(ii) In Î”DBO,

OD = OB (radii of the same circle)

∴ ∠OBD = ∠BDO = ∠CBD (from (ii))

∠ABD = ∠CBD

AD = CD (∵ Equal chords subtend equal angles)

17. (a) In the figure, (i) given below AB and XY are diameters of a circle with centre O. If APX = 30°. Find

(i) AOX

(ii) APY

(iii) BPY

(iv) OAX.

(b) In the figure (ii) given below, AP and BP are tangents to the circle with centre O. If CBP = 25° and CAP = 40°, find :

(ii) AOB

(iii) ACB

(iv) APB

(a) AB and XY are diameters of a circle with centre O.

∠APX = 30°

To find :

(i) ∠AOX

(ii) ∠APY

(iii) ∠BPY

(iv) ∠OAX

(i) ∵ arc AX subtends ∠AOX at the centre and ∠APX at the remaining point of the circle.

∴ ∠AOX = 2 ∠APX

= 2 × 30°

= 60°

(ii) ∵ XY is the diameter

∴ ∠XPY = 90° (Angle in a semi-circle)

∴ ∠APY = ∠XPY - ∠APX

= 90° - 30°

= 60°

(iii) ∠APB = 90° (Angle in a semi-circle)

∠BPY = ∠APB - ∠APY

= 90° - 60°

= 30°

(iv) In Î”AOX,

OA = OX (radii of the same circle)

∴ ∠OAX = ∠OXA

But ∠AOX + ∠OAX + ∠OXA = 180° (Angles of a triangle)

⇒ 60° + ∠OAX + ∠OAX = 180°

⇒ 2∠OAX = 180°- 60°

= 120 ̊

∴ ∠OAX = 120°/2

= 60°

(b) Join CD

(i) ∠CDB = ∠CBP (Angles in the alternate segments)

∴ ∠CDB = 25° …(i)

Similarly, ∠CDA = ∠CAP = 40° (Angles in the alternate segments)

∴ ∠ADB = ∠CDA + ∠CDB

= 40° + 25°

= 65°

(ii) arc ACB subtends ∠AOB at the centre and ∠ADB at the remaining part of the circle.

∴ ∠AOB = 2∠ADB

= 2 × 65°

= 130°

(iii) ACBD is a cyclic quadrilateral

∴ ∠ACB + ∠ADB = 180°

⇒ ∠ACB + 65° = 180°

⇒ ∠ACB = 180° - 65°

= 115°

(iv) ∠AOB + ∠APB = 180°

⇒ 130° + ∠APB = 180°

⇒ ∠APB = 180° - 130°

= 50°

The solutions provided for Chapter 15 Circles of ML Aggarwal Textbook. This solutions of ML Aggarwal Textbook of Chapter 15 Circles contains answers to all the exercises given in the chapter. These solutions are very important if you are a student of ICSE boards studying in Class 10. While preparing the solutions, we kept this in our mind that these should based on the latest syllabus given by ICSE Board.

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