ML Aggarwal Solutions for Chapter 15 Circles Class 10 Maths ICSE
Exercise 15.1
1. Using the given information, find the value of x in each of the following figures:
(i)
(i) ∠ADB and ∠ACB are in the same segment.
∠ADB = ∠ACB = 50°
Now in ∆ADB,
∠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
∠BAD = ∠BCD
Because angles in the same segment
∠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)
(iii)
(iv)
(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° + x° = 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 × 70o = 35o
∵ ∠CMO = 90o
∴ ∠AMC = 90o
(∴∠AMC + ∠CMO = 180o)
∠ACM + ∠AMC + ∠CAM = 180o
⇒ 35o + 90o + xo = 180o
⇒ 125o + xo = 180o
⇒ xo = 180 – 125o = 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)
Answer
(a) Construction: Join AB
∠A = ∠C = 35° (Alt Angles)
⇒ ∠ABC = 35o
(b) ∠AOC + reflex ∠AOC = 360o
⇒ 130o + Reflex ∠AOC = 360o
⇒ Reflex ∠AOC = 360o – 130o = 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 × 230o = 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)
(a) ABCD is cyclic Quadrilateral
∠B + ∠D = 180°
⇒ y + 40° + 45o = 180o
(y + 85o = 180o)
⇒ y = 180o – 85o = 95o
(∠ACB = ∠ADB)
⇒ xo = 40
(b) Arc ADC Subtends ∠AOC at the center and
∠ ABC at the remaining part of the circle
∠AOC = 2 ∠ABC
⇒ xo = 60o
Again, ABCD is a Cyclic quadrilateral
∠B + ∠D = 180o
(60o + yo = 180o)
⇒ y = 180o – 60o = 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°.
∠NYB + ∠YNB + ∠YBN = 180o
⇒ 50o + 20o + ∠YBN = 180o
⇒ ∠YBN + 70o = 180o
⇒ ∠YBN = 180o – 70o = 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 = 2×110o =220o
(b) (i) ∠AOB + reflex ∠AOB = 360o
(Angles at the point)
⇒ 140o + reflex ∠AOB = 360o
⇒ Reflex ∠AOB = 360o – 140o = 220o
⇒ 50o + 110o + 140o + ∠OBC =360°
⇒ 300o + ∠OBC = 360°
⇒ ∠300o + ∠OBC = 360°
⇒ ∠OBC = 360o – 300o
⇒ ∠OBC = 60o
(ii) In Quadrilateral, OACB
∠OAC + ∠ACB + ∠AOB + ∠OBC = 360o
⇒ 50o + 110o + 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 – 140o = 40o
⇒ ∠OAB = 40o/2 = 20°
But, ∠OBC = 60o
∠CBA = ∠OBC – ∠OBA = 60o – 20o = 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
= 2×32o = 2 ∠CDE
= 2×32o
= 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)
Answer
(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)
Now in △ADC,
∠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.
(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 × 130o = 65o
But ∠ACB + ∠BCD = 180o (Linear Pair)
⇒ 65o + ∠BCD = 180o
⇒ ∠BCD = 180o – 65o = 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 = 360o (Angles at a point)
⇒ 230o + x = 360o
⇒ x = 360o – 230o = 130o
(b) Join OC
In ∆ABC,
AC = BC
∠A = ∠B
But ∠A + ∠B + ∠C = 180o
⇒ ∠A + ∠A + 56° = 180°
⇒ 2∠A = 180o – 56° = 124o
⇒ ∠A = 124/2 = 62o or ∠CAB = 62°OC is the radius of the circle. OC bisects ∠ACB.
∠OCA = ½∠ACB = ½×56o = 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)
Answer
From the figure
(i) ABCD is a cyclic quadrilateral
Ext. ∠DCE = ∠BAD
∠BAD = xo
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 = 180o (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
xo = 2×100o = 200o
(iii) In ∆ACB,
∠CAB + ∠ABC + ∠ACB = 180o (Angles of a triangle)
But ∠ACB = 90o (Angles of a semicircle)
25o + 90o + ∠ABC = 180o
⇒ 115o + ∠ABC = 180o
⇒ ∠ABC = 180o – 115° =65o
ABCD is a cyclic quadrilateral
∠ABC + ∠ADC = 180o (Opposite angles of a cyclic quadrilateral)
⇒ 65o + xo =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.
Answer
(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 = 180o (Sum of opposite angels in a cyclic quadrilateral is 180o)
⇒ 75o + ∠ADC = 180o
⇒ ∠ADC + 180o – 75o
⇒ ∠ADC = 105o
(b) (i) AC is the diameter of the circle
∠ABC = 90o (Angle in a semi-circle)
(ii) ABCD is a cyclic quadrilateral
∠BAD + ∠BCD = 180o
⇒ ∠BAD + 75o = 180o (∠BCD = 75o)
⇒ ∠BAD = 180o - 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
(i) ∠CAD (ii) ∠CBD (iii) ∠ADC
(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 – 32o = 148o (opposite angles of cyclic quadrilateral)
(iii) ∠BAC = ∠BDC = 32o (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 = 25o (∠BAC = 25o)
Now AB ∥ DC and BD is the transversal
∠BDC = ∠ABD
⇒ ∠ABD = 25o
⇒ ∠ABC = ∠ABD + ∠CBD = 25o + 55o = 80o
But ∠ABC + ∠ADC = 180o (opposite angles of a cyclic quadrilateral)
⇒ 80o + ∠ADC = 180o
⇒ ∠ADC = 180o – 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.
(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 = 180o (Sum of opposite angles = 180o)
⇒ ∠ABC + 80o = 180o
∠AOE = 150o, ∠DAO = 51o
To find ∠BEC and ∠EBC
ABED is a cyclic quadrilateral
Ext. ∠BEC = ∠DAB = 51o
⇒ ∠AOE = 150o
⇒ Ref. ∠AOE = 360o – 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 ×210o = 105o
But Ext. ∠EBC = ∠ADE = 105o
Hence, ∠BEC = 51o and ∠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)
(a) ADFE is a cyclic quadrilateral
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°
∠BAD = 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.
(i) ∠QBC
(ii) ∠BCP
Answer
(i) ABCD is a cyclic quadrilateral
⇒ 30o + p = 180o
⇒ p = 180o – 30o = 150o
(ii) Arc BD subtends ∠BOD at the center
And ∠BAD at the remaining part of the circle
∠BOD = 2∠BAD
⇒ q = 2×30o = 60o
∠BAD = ∠BED are in the same segment of the circle
⇒ ∠BAD = ∠BED
⇒ 30o = 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
(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 = 32o (Alternate angles)
Now, PRST is a cyclic quadrilateral,
∠STP + ∠SRP = 180o
⇒ ∠STP = 180o – 32o = 148o
(b) In the given figure,
∠ACE 43o and ∠CAF = 620
Now, in ∆AEC
∠ACE + ∠CAE + ∠AEC = 180o
⇒ 43o + 62o + ∠AEC = 180o
⇒ 105o + ∠AEC = 180o
⇒ ∠AEC = 180o – 105° = 75o
But ∠ABD + ∠AED = 180° (sum of opposite angles of acyclic quadrilateral) and,
∠AED = ∠AEC
⇒ a + 75o = 180o
⇒ a = 180o – 75o – 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 = 180o (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.
Answer
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,
OP2 = OT2 + PT2
⇒ (5)2 = (3)2 + PT2
⇒ PT2 = (5)2 – (3)2 = 25 – 9 = 16 = (4)2
⇒ PT = 4 cm.
2. A 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.
Answer
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)2 = 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,
OP2 = OA2 + AP2 (By Pythagoras axiom)
= (6)2 + (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.
Answer
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.
Answer
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
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.
Answer
(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.
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)
OB2 = OA2 – AB2
⇒ r2 = (r + 7.5)2 – 152
⇒ r2 = r2 + 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.
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
⇒ OP2 = 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.
Answer
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.
Answer
(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 ΔPQR.
(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.
Answer
(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 ΔPQR ∠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 ΔOAP
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
(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 ΔADO')/(area of ΔACO)
(a) Let x be the radius of the circle with centre C and radii of each equal
CP = x + 2 and CM = 4 – x
∴ In right ΔPCM
(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.
∠A = ∠A (Common)
∠D = ∠C (each 90°)
∴ ΔADO' ~ ΔACO (AA axiom of similarity)
(i) ∴ AO'/AO = r/3r = 1/3
(ii) ∵ ΔADO' ~ ΔACO
∴ (area of ΔADO')/(area of ΔACO)
= (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.
Answer
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.
Answer
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 ΔPAB,
PB2 = PA2 + AB2
= 62 + 82
= 36 + 64
= 100
= 102
∴ PB = 10 cm
Now in ΔPAB and ΔBCQ,
∴ ∠A = ∠C (each 90°)
∠ABP = ∠CBQ (vertically opposite angles)
∴ ΔPAB ~ ∠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
Answer
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 ΔATP and ΔSBP
∠T = ∠S (each 90°)
∠APT = ∠BPS (vertically opposite angles)
∴ ΔATP ~ ΔSBP (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 ΔATP,
AP2 = AT2 + TP2
⇒ (10)2 = (6)2 + TP2
⇒ 100 = 36 + TP2
⇒ TP2 = 100 – 36 = 64
⇒ TP2 = (8)2
∴ TP = 8 cm
Similarly in right ΔPSB,
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 ΔOAB,
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 ΔABD
∠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 ΔQRS,
∠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.
(a) PQ is tangent and Ad is chord
(i) ∴ ∠QAB = ∠BDA = 30° (Angles in the alternate segment)
(ii) In ΔADB,
∠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 ΔBCD
∠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 ΔBCF,
∠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 ΔTSR,
∠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 ΔOSP,
∠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.
Answer
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°.
Answer
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
(i) ∠AOB
(ii) ∠OAB
(iii) ∠ACB.
(a) AB is the diameter.
∴ ∠ACB = 90° and ∠CAB = 34°
∴ In ΔABC,
∠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 ΔOAB,
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 ΔOPQ,
∠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.
Answer
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.
(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 ΔOTP,
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.
Answer
∴ 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 ΔACD ~ ΔDCB.
(ii) Find the length of CD.
In ΔACD and ΔDCB
∠C = ∠C (common)
∠CAD = ∠CDB [Angle between chord and tangent is equal to angle made by chord in alternate segment.]
∴ ΔACD ~ ΔDCB
∴ 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.
Answer
(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)
ΔABC 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.
Answer
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.
(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 ΔABC,
∠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 ΔATC,
Ext.∠CAB = ∠ATC + ∠TCA
= 36° + 48°
= 84°
But ∠TCA = ∠ABC (Angles in the alternate segment)
∴ ∠ABC = 48°
But in ΔABC,
∠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 ΔABC is isosceles with AB = AC. Prove that the tangent at A to the circumcircle of ΔABC is parallel to BC.
Given : ΔABC 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 ΔABC,
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.
Answer
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
Adding, we get
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.
(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 ΔACB,
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.
(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 ΔOAB,
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 ΔADB, ΔADB = 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 ΔOAB,
∠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°
Answer
(b) 50°
ABCD is a cyclic quadrilateral,
AB is the diameter of the circle circumscribing it
∠ADC = 140°, ∠BAC = Join AC
140° + ∠ABC = 180°
∠ABC = 180° - 140° = 40°
Now in ΔABC,
∠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°
Answer
(c) 60°
In the given figure, O is the centre of the circle ∠BAO = 60°
In ΔABO, 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 ΔAOB,
∠AOB = 90°, ∠ABC = 30°
In ΔAOB, ∠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 ΔACB, ∠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
ΔOPQ 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 ΔOAB,
∠OAB = ∠OBA = 3x (OA = OB)
Sum of angles of ΔOAB = 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
Answer
(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 ΔOPQ
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
Answer
(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 ΔOPQ = 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°
Answer
(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 ΔOPR
∠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°
Answer
(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
Answer
(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 ΔOEC,
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
Answer
(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 ΔOAC
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°
Answer
(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°
Answer
(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°
Answer
(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 ΔOAB,
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 ΔABC 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 ΔABC 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 ΔABC 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 ΔABC 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 ΔABC 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 ΔABC = 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 ΔOPQ is an isosceles triangle, then ∠QOP is equal to
(a) 30°
(b) 60°
(c) 45°
(d) 90°
Answer
(c) 45°
PQ is tangent to the circle at point P centre of the circle is O.
ΔOPQ is an isosceles triangle
OP = PQ
∵ OP is radius and PQ is tangent to the circle
∴ OP ⊥ PQ i.e., ∠OPQ = 90°
∵ OP = PQ (∵ ΔOPQ 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 ΔOPT
∠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°
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 ΔAPC and ΔBPD
∠P = ∠P (common)
∠A = ∠BDP {Ext. of a cyclic quad. is equal to its interior opposite angles}
∴ ΔAPC ~ ΔBPD (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
⇒ 60o + ∠E = 180o
⇒ 600 + ∠E = 180o (∠E = 180o – 60o)
⇒ ∠E = 120o
Hence, ∠BDC = 60o and ∠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)
∠OAD = ∠ODA
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
Solution:
(a) (i) Join DB, CA and CB. ∠ADC = 118° (given) and ∠ADB = 90° (Angles in a semi-circle)
∠BDC = ∠ADC – ∠ADB
= 118° – 90o = 28o (ABCD is a cyclic quadrilateral)
(ii) ∠ADC + ∠ABC = 180o
⇒ 118o + ∠ABC = 180o
⇒ ∠ABC = 180o – 118o = 62o
But in ∆AEB,
∠AEB = 90o (Angles in a semi-circle)
∠EAB = ∠ABE (AE = BE)
∠EAB + ∠ABE = 90o
⇒ ∠EAB = 90o × ½ = 45o
⇒ ∠CBE = ∠ABC + ∠ABE
= 62o + 45o = 107o
But AEBD is a cyclic quadrilateral
∠CAE + ∠CBE = 180o
⇒ ∠CAE + 107o = 180o
⇒ ∠CAE = 180o – 107o = 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 = 90o (Angles in a semi circle)
⇒ ∠AED = ∠AEB + ∠BED
= 90o + 40o = 130o
(ii) Now in cyclic quadrilateral AEDB
∠AED + ∠DBA = 180o
⇒ 130o + ∠DBA =180o
⇒ ∠BDA = 180o – 130o = 50o
Chord AE = ED (given)
∠DBE = ∠EBA
But ∠DBE + ∠EBA = 50o
⇒ DBE + ∠DBE = 50o
⇒ 2∠DBE = 50o
⇒ ∠DBE = 25o or ∠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
Answer
(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 = 180o – (∠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 {180o – (ACB + BAC)}
But ∠AOC = 360o – 2(∠ACB + ∠BAC)
But ∠AOC = 360o – reflex ∠AOC
= 360 ̊ – (360o – 2(∠ACB + ∠BAC)
= 360o – 360o + 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:
(i) ∠ADC
(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
(iv) ∠ADB
(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 ΔACD,
∠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 ΔABD and ΔCBD
AB = CD (given)
BD = BD (Common)
∠BAD = ∠BCD (Angles in the same segment)
∴ ΔABC ≅ ΔCBD (SSA axiom of congruency)
∴ BC = AD
(b) Given: ΔABC is an isosceles triangle and ∠ABC = 50°
In ΔABC, an isosceles triangle
⇒ ∠ACB = ∠ABC (∵ opp. ∠s of an isosceles Δs)
⇒ ∠ACB = 50°
Also in ΔABC
⇒ ∠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.
Answer
Join OT, OP = 13 cm and TP = 12 cm
∵ OT is the radius
∴ OT ⊥ TP
Now in right ΔOTP
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.
Answer
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.
Answer
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 ΔOAP and OBP
Hypotenuse OP = OP (common)
Side OA = OB (radii of the same circle)
∴ ΔOAP ≅ ΔOBP (R.H.S axiom of congruency)
∴ ∠AOP = ∠BOP (C.P.C.T.)
And ∠APO = ∠BPO (C.P.C.T)
Now in ΔAPM and ΔBPM,
PM = PM (common)
∠APM = ∠BPM (proved)
AP = BP (tangents from P to the circle)
∴ ΔAPM ≅ ΔBPM (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.
(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 ΔAPC and ΔBQC
∠PCA = ∠QCB (vertically opposite angles)
∠APC = ∠BQC (each 90°)
∴ ΔAPC ~ ΔBQC
∴ 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 ΔOAQ
OQ = OA (Radii of the same circle)
∴ ∠OQA = ∠POB (Corresponding angles)
And ∠OQA = ∠QOP (Alternate angles)
But ∠QAO = ∠OQA (Proved)
∠POB = ∠QOB
Now, in ΔOPQ and ΔOBP
OP = OP (Common)
OQ = OB (Radii of the same circle)
∠BOP = ∠POB (Proved)
∴ ΔOPQ ≅ ΔOBP (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 ΔAPM,
AP2 = AM2 + PM2
⇒ 172 = AM2 + 152
⇒ AM2 = 172 – 152
= 289 – 225
= 64
= (8)2
∴ AM = 8 cm
Similarly in right ΔBNQ
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.
Answer
∵ 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.
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
In ΔADP,
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
(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 ΔABC 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 ΔBCD, 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 ΔAOB
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 ΔDOE,
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 ΔDBO,
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.
(i) ∠ADB
(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 ΔAOX,
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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