# ML Aggarwal Solutions for Chapter 15 Circle Class 9 Maths ICSE

**Exercise 15.1**

**1. Calculate the length of a chord which is at a distance of 12 cm from the centre of a circle of radius 13 cm.**

**Solution**

AB is chord of a circle with center O and OA is its radius OM ⊥ AB

Therefore, OA = 13 cm, OM = 12 cm

Now from right angled triangle OAM,

OA^{2} = OM^{2} + AM^{2} by using Pythagoras theorem,

⇒ 13^{2} = 12^{2} + AM^{2}

⇒ AM^{2} = 13^{2} – 12^{2}

⇒ AM^{2} = 169 – 144

⇒ AM^{2} = 25

⇒ AM^{2} = 5^{2}

⇒ AM = 5

We know that OM perpendicular to AB

Therefore, M is the midpoint of AB

AB = 2 AM

⇒ AB = 2 (5)

⇒ AB = 10 cm

**2. A chord of length 48 cm is drawn in a circle of radius 25 cm. Calculate its distance from the center of the circle.**

**Solution**

AB is the chord of the circle with centre O and radius OA

OM is perpendicular to AB

Therefore, AB = 48 cm

OA = 25 cm

OM ⊥ AB

M is the mid-point of AB

AM = 1/2 AB = ½ × 48 = 24 cm

Now right ∆OAM,

OA^{2 }= OM^{2 }+ AM^{2 }**(by Pythagoras Axiom)**

⇒ (25)^{2} = OM ^{2 }+ (24)^{2}

⇒ OM^{2 }= (25)^{2 }– (24)^{2 }= 625 – 576

⇒ OM^{2 }= 49 = (7)^{2}

⇒ OM = 7 cm

**3. A chord of length 8 cm is at a distance of 3 cm from the centre of the circle. Calculate the radius of the circle.**

**Solution**

AB is the chord of a circle with center O

And radius OA and OM ⊥ AB

AB = 8 cm

OM = 3 cm

OM ⊥ AB

M is the mid-point of AB

AM = ½ AB = ½ × 8 = 4 cm.

Now in right ∆OAM

OA^{2 }= OM^{2 }+ AM^{2 }**(By Pythagoras Axiom)**

= (3)^{2 }+ (4)^{2 }= 9 + 16 = 25

= (5)^{2}

⇒ OA = 5 cm.

**4. Calculate the length of the chord which is at a distance of 6 cm from the centre of a circle of diameter 20 cm.**

**Solution**

AB is the chord of the circle with centre O

And radius OA and OM ⊥ AB

Diameter of the circle = 20 cm

Radius = 20/2 = 10 cm

OA = 10 cm, OM = 6 cm

Now in right ∆OAM,

OA^{2 }= AM^{2 }+ OM^{2 }**(By Pythagoras Axiom)**

⇒ (10)^{2} = AM^{2 }+ (6)^{2}

⇒ AM^{2} = 10^{2} – 6^{2}

⇒ AM^{2 }= 100 – 36 = 64 = (8)^{2}

⇒ AM = 8 cm

OM ⊥ AB

M is the mid-point of AB.

⇒ AB = 2× AM = 2×8 = 16 cm.

**5. A chord of length 16 cm is at a distance of 6 cm from the centre of the circle. Find the length of the chord of the same circle which is at a distance of 8 cm from the centre.**

**Solution**

AB is a chord a circle with centre O and

OA is the radius of the circle and OM ⊥ AB

AB = 16 cm, OM = 6 cm

OM ⊥ AB

AM = ½ AB = ½ × 16 = 8 cm

Now in right ∆OAM

OA^{2 }= AM^{2 }+ OM^{2 }**(By Pythagoras Axiom)**

= (8)^{2} + (6)^{2}

⇒ 64 + 36 = 100 = (10)^{2}

Now CD is another chord of the same circle

ON ⊥ CD and OC is the radius.

In right ∆ONC

OC^{2 }= ON^{2 }+ NC^{2 }**(By Pythagoras Axioms)**

⇒ (10)^{2 }= (8)^{2 }+ (NC)^{2}

⇒ 100 = 64 + NC^{2}

⇒ NC^{ 2 }= 100 – 64 = 36 = (6)^{2}

⇒ NC = 6

But ON ⊥ AB

N is the mid-point of CD

⇒ CD = 2 NC = 2×6 = 12 cm

**6. In a circle of radius 5 cm, AB and CD are two parallel chords of length 8 cm and 6 cm respectively. Calculate the distance between the chords if they are on :**

**(i) the same side of the centre.**

**(ii) the opposite sides of the centre**

**Solution**

Two chords AB and CD of a Circle with centre O and radius OA or OC

OA = OC = 5 cm

AB = 8 cm

CD = 6 cm

OM and ON are perpendiculars from O to AB and CD respectively.

M and N are the Mid-points of AB and

CD respectively

In figure (i) chord are on the same side

And in figure (ii) chord are on the opposite

Sides of the centre

In right ∆OAM

OA^{2} = AM^{2 }+ OM^{2 }**(By Pythagoras Axiom)**

⇒ (5)^{2} = (4)^{2} + OM^{2}

⇒ AM = ½ AB

⇒ 25 = 16 + OM^{2}

⇒ OM^{2} = 25 – 16 = 9= (3)^{2}

⇒ OM = 3 cm

Again in right ∆OCN,

OC^{2 }= CN^{2 }+ ON^{2}

⇒ (5)^{2 }= (3)^{2 }+ ON^{2}

⇒ (CN = ½ CD)

⇒ 25 = 9 + ON^{2}

⇒ ON^{2} = 25 – 9 = 16 = (4)^{2}

⇒ ON = 4

In fig (i), distance MN = ON – OM

= 4 – 3 = 1cm.

In fig (ii)

MN = OM + ON = 3 + 4 = 7 Cm

**7. (a) In the figure given below, O is the centre of the circle. AB and CD are two chords of the circle, OM is perpendicular to AB and ON is perpendicular to CD. AB = 24 cm, OM = 5 cm, ON = 12 cm. Find the:**

**(i) radius of the circle.**

**(ii) length of chord CD.**

**(b) In the figure (ii) given below, CD is the diameter which meets the chord AB in E such that AE = BE = 4 cm. If CE = 3 cm, find the radius of the circle.**

**Solution**

(a)Given : AB = 24 cm, OM = 5cm, ON = 12cm

OM ⊥ AB

M is midpoint of AB

AM = 12 cm

**(i)** Radius of circle OA = √OM^{2 }+ AM^{2}

**(ii)** Again OC^{2 }= ON^{2 }+ CN^{2}

⇒ 13^{2 }= 12^{2} + CN^{2}

⇒ CN = √(13^{2} – 12^{2}) = √(169 – 144) = √25

⇒ CN = 5 cm

As ON ⊥ CD, N is mid-Point of CD

CD = 2CN = 2×5 = 10 cm

**(b)** AB = 8 cm, EC = 3 cm

Let radius OB = OC = r

OE = (r-3) cm

Now in right ∆OBE

OB^{2 }= BE^{2 }+ OE^{2}

⇒ r^{2 }= (4)^{2} + (r – 3)^{2}

⇒ r^{2 }= 16 + r^{2 }– 6r + 9

⇒ 6r = 25

⇒ r = 25/6 = 4 1/6 cm

**8. In the adjoining figure, AB and CD ate two parallel chords and O is the centre. If the radius of the circle is 15 cm, find the distance MN between the two chords of length 24 cm and 18 cm respectively.**

**Solution**

In the figure, chords AB ∥ CD

O is the centre of the circle

Radius of the Circle = 15 cm

Length of AB = 24 cm and CD = 18 cm

Join OA and OC

AB = 24 cm and OM ⊥ AB

AM = MB = 24/2 = 12 cm

Similarly ON ⊥ CD

CN = ND = 18/2 = 9 cm

Similarly In right ∆ CNO

OC^{2 }= CN^{2} + ON^{2 }(15)^{2} = (9)^{2 }+ ON^{2}

⇒ 225 = 81 + ON^{2}

⇒ ON^{2 }= 225 – 81 = 144 = (12)^{2}

⇒ ON = 12 cm

Now MN = OM + ON = 9 + 12 = 21 cm

**9. AB and CD are two parallel chords of a circle of lengths 10 cm and 4 cm respectively. If the chords lie on the same side of the centre and the distance between them is 3 cm, find the diameter of the circle.**

**Solution **

AB and CD are two parallel chords and AB = 10 cm, CD = 4 cm and distance between

AB and CD = 3 cm

Let radius of circle OA = OC = r

OM ⊥ CD which intersects AB in L.

Let OL =x, then OM = x + 3

Now right ∆OLA

OA^{2} = AL^{2 }+ OL^{2}

⇒ r^{2 }= (5)^{2 }+ x^{2 }= 25 + x^{2 }**(l is mid- point of AB)**

Again in right ∆OCM

OC^{2 }= CM^{2} + OM^{2}

⇒ r^{2 }= (2)^{2 }+ (x + 3)^{2 }**(M is mid-point of CD)**

⇒ r^{2 }= 4 + (x + 3)^{2 }**(M is mid-Point of CD)**

⇒ r^{2 }= 4 + (x + 3)^{2}

from (i) and (ii)

25 + x^{2 }= 4 + (x + 3)^{2}

⇒ 25 + x^{2 }= 4 + x^{2 }+ 9 + 6x

⇒ 6x = 25 – 13 = 12

⇒ x = 12/6 = 2 cm

Substituting the value of x in (i)

r^{2 }= 25 + x^{2} = 25 + (2)^{2} = 25 + 4

⇒ r^{2 }= 29

⇒ r = √29cm

Diameter of the circle = 2 r

= 2 ×√29 cm = 2√29 cm

**10. ABC is an isosceles triangle inscribed in a circle. If AB = AC = 12√5 cm and BC = 24 cm, find the radius of the circle.**

**Solution **

AB = AC 12√5 and BC = 24 cm.

Join OB and OC and OA

Draw AD ⊥ BC which will pass through

Centre O

OD bisect BC in D

BD = DC = 12 cm

In right ∆ABD

AB^{2 }= AD^{2 }+ BD^{2}

⇒ (12√5)^{2 }= AD^{2 }+ BD^{2}

⇒ (12√5)^{2 }= AD^{2} + (12)^{2}

⇒ 144×5 =AD^{2} + 144

⇒ 720 – 144 = AD^{2}

⇒ AD^{2 }= 576

⇒ AD = √576 = 24

Let radius of the circle = OA = OB = OC = r

OD = AD – AO = 24 – r

Now in right ∆OBD,

OB^{2} = BD^{2} + OD^{2}

⇒ r^{2 }= (12)^{2 }+ (24 – r)^{2}

⇒ r^{2 }=144 + 576 + r^{2 }– 48r

⇒ 48r = 720

⇒ 48r = 720

⇒ r = 720/48 = 48r

⇒ 48r = 720

⇒ r = 720/48 = 15cm.

Radius = 15 cm.

**11. An equilateral triangle of side 6 cm is inscribed in a circle. Find the radius of the circle.**

**Solution **

ABC is an equilateral triangle inscribed in a

Circle with centre O. Join OB and OC,

From A, Draw AD ⊥ BC which will pass

Through the centre O of the circle.

Each side of ∆ABC = 6 cm.

AD = √3/2 a= √3/2 × 6 = 3 √3 cm.

OD = AD – AO = 3√3 – r

Now in right ∆OBD

OB^{2 }= BD^{2 }+ OD^{2}

⇒ r^{2 }= (3)^{2 }+ (3√3-r)^{2}

⇒ r^{2 }= 9 + 27 + r^{2 }-6√3r **(D is mid-point of BC)**

⇒ 6√3r = 36

⇒ R = 36/6√3 = 6/√3 × √3/√3 = 6√3/3 = 2√3 cm

Radius = 2√3 cm

**12. AB is a diameter of a circle. M is a point in AB such that AM = 18 cm and MB = 8 cm. Find the length of the shortest chord through M.**

**Solution**

AM = 18 cm and MB = 8 cm

AB = AM + MB = 18 + 8 = 26 cm

Radius of the circle = 26/2 = 13 cm

Let CD is the shortest chord drawn through M.

CD ⊥ AB

Join OC

OM = AM – AO = 18 – 13 = 5 cm

OC = OA = 13 cm

Now in right ∆OMC

OC^{2 }= OM^{2} + MC^{2}

⇒ (13)^{2} = (5)^{2 }+ MC^{2 }(MC^{2 }= 13^{2 }– 5^{2}

⇒ MC^{2 }= 169 – 25 = 144 = (12)^{2}

⇒ MC = 12

M is Mid-Point of CD

CD = 2 ×MC = 2×12 = 24 cm

**Exercise 15.2**

**1. If arcs APB and CQD of a circle are congruent, then find the ratio of AB: CD.**

**Solution**

arc APB = arc CQD **(given)**

AB = CD **(If two arcs are congruent, then their corresponding chords are equal)**

Ratio of AB and CD = AB / CD = AB /AB = 1/1

AB : CD = 1 : 1

**2. A and B are points on a circle with centre O. C is a point on the circle such that OC bisects ∠AOB, prove that OC bisects the arc AB.**

**Solution**

Given : in a given circle with centre O,A

And B are Two points on the circle. C i

another point on the circle such that

∠AOC = ∠BOC

To prove : arc AC = arc BC

**Proof:** OC is the bisector of ∠AOB

⇒ ∠AOC = ∠BOC

But these are the angle subtended by the arc AC and BC

arc AC = arc BC.

**3. Prove that the angle subtended at the centre of a circle is bisected by the radius passing through the mid-point of the arc.**

**Solution **

Given: AB is the arc of the circle with

Centre O and C is the mid-Point od arc AB.

To prove: OC bisects the ∠AOB

I,e ∠AOC = ∠BOC

**Proof:**

C is the mid-point of arc AB.

arc AC = arc BC

But arc AC and arc BC subtend ∠AOC and

∠BOC at the centre

∠AOC = ∠BOC

Hence, OC Bisects the ∠AOB.

**4. In the given figure, two chords AB and CD of a circle intersect at P. If AB = CD, prove that arc AD = arc CB.**

**Solution **

Given: two chord AB and CD of a Circle

Intersect at P and AB = CD

To prove : arc AD = arc CB

**Proof:**

AB = CD **(given)**

minor arc AB = minor arc CD

Subtracting arc BD from both sides

arc AB = arc BD = arc CD – arc BD

⇒ arc AD = arc CD

**Chapter test**

**1. In the given figure, a chord PQ of a circle with centre O and radius 15 cm is bisected at M by a diameter AB. If OM = 9 cm, find the lengths of :** **(i) PQ** **(ii) AP** **(iii) BP**

**Solution**

Given, radius = 15 cm

OA = OB = OP = OQ = 15 cm

Also, OM = 9 cm

MB = OB – OM = 15 – 9 = 6 cm

AM = OA + OM =15 + 9 cm = 24 cm

In ∆OMP,

By using Pythagoras Theorem,

OP^{2 }= OM ^{2 }+ PM^{2}

⇒ 15^{2} = 9^{2 }+ PM^{2}

⇒ PM^{2} = 255 – 81

⇒ PM = √144 = 12 cm

Also, In ∆OMQ

By using Pythagoras Theorem

OQ^{2} = OM^{2 }+ QM^{2}

⇒ 15^{2 }= OM^{2} + QM^{2}

⇒ 15^{2 }= 9^{2} + QM^{2 }**(QM ^{2 }= 225 – 81)**

⇒ QM = √144 = 12 cm

⇒ PQ = PM + QM **(As radius is bisected at M)**

⇒ PQ = 12 + 12 cm = 24 cm

**(ii) **Now in ∆APM

AP^{2 }= AM^{2 }+ OM^{2}

⇒ AP^{2 }=24^{2} + 12^{2}

⇒ AP^{2 }= 576 + 144

⇒ AP = √720 = 12 √5 cm

**(iii) **Now in ∆BMP

BP^{2} = BM^{2 }+ PM^{2}

BP^{2 }= 6^{2 }+ 12^{2}

⇒ BP^{2 }= 36 + 144

⇒ BP = √180 = 6√5 cm

**2. The radii of two concentric circles are 17 cm and 10 cm ; a line PQRS cuts the larger circle at P and S and the smaller circle at Q and R. If QR = 12 cm, calculate PQ.**

**Solution **

A line PQRS intersects the outer circle at P

And S and inner circle at Q and R radius of

Outer circle OP = 17 cm and radius of inner

Circle OQ = 10 cm

QR = 12 cm

From O, draw OM ⊥ PS

QM = ½ QR = ½ × 12 = 6 cm

In right ∆OQM

OQ^{2} = OM^{2} + QM^{2}

⇒ (10)^{2 }= OM^{2 }+ (6)^{2}

⇒ OM^{2 }= 10^{2} – 6^{2}

⇒ OM^{2 }= 100 – 36 = 64 = (8)^{2}

⇒ OM = 8 cm

Now in right ∆OPM

OP^{2} OM^{2} + PM^{2}

⇒ (17)^{2 }= OM^{2 }+ PM^{2}

⇒ PM^{2 }= (17)^{2} – (8)^{2}

⇒ PM^{2 }= 289 – 64 = 225 = (15)^{2}

⇒ PM = 15 cm

⇒ PQ = PM – QM = 15 – 6 = 9 cm