ML Aggarwal Solutions For Class 9 Maths Chapter 15 Circle is one of the best ways to strengthen oneâ€™s skills and knowledge. It contains all the relevant study material that can help the students score well in the examinations. ML Aggarwal Solutions Class 9 Chapter 15 Circle is an important chapter from the examination perspective. A circle is a special kind of ellipse in which the eccentricity is zero and the two foci are coincident. A circle is also termed as the locus of the points drawn at an equidistant from the centre. The distance from the center of the circle to the outer line is its radius. Diameter is the line which divides the circle into two equal parts and is also equal to twice the radius. A proper understanding of this chapter will further help the students with a few other chapters in higher classes as well. The solutions are provided by the subject experts and are accurate. Every question is explained stepwise for a better understanding of the students. They can use these for reference purposes. Chapter 15 of ML Aggarwal Solutions for Class 9 Maths explains circle, types of circle and properties of circles.

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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 âŠ¥ 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} â€“ 12^{2}

AM^{2} = 169 â€“ 144

AM^{2} = 25

AM = 5^{2}

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 âŠ¥ AB

M is the mid-point of AB

AM = 1/2 AB = Â½ Ã— 48 = 24 cm

Now right âˆ†OAM,

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

(by Pythagoras Axiom)

(25)^{2} = OM ^{2 }+ (24)^{2}

OM^{2 }= (25)^{2 }â€“ (24)^{2 }= 625 â€“ 576

= 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 âŠ¥ AB

AB = 8 cm

OM = 3 cm

OM âŠ¥ AB

M is the mid-point of AB

AM = Â½ AB = Â½ Ã— 8 = 4 cm.

Now in right âˆ†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 âŠ¥ AB

Diameter of the circle = 20 cm

Radius = 20/2 = 10 cm

OA = 10 cm, OM = 6 cm

Now in right âˆ†OAM,

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

(By Pythagoras Axiom)

(10)^{2} = AM^{2 }+ (6)^{2}

AM^{2} = 10^{2} â€“ 6^{2}

AM^{2 } = 100 â€“ 36 = 64 = (8)^{2}

AM = 8 cm

OM âŠ¥ AB

M is the mid-point of AB.

AB = 2 AM = 2 Ã— 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 âŠ¥ AB

AB = 16 cm, OM = 6 cm

OM âŠ¥ AB

AM = Â½ AB = Â½ Ã— 16 = 8 cm

Now in right âˆ†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 âŠ¥ CD and OC is the radius.

In right âˆ†ONC

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

(By Pythagoras Axioms)

(10)^{2 }= (8)^{2 }+ (NC)^{2}

100 = 64 + NC^{2}

NC^{ 2 }= 100 â€“ 64 = 36 = (6)^{2}

NC = 6

But ON âŠ¥ AB

N is the mid-point of CD

CD = 2 NC = 2 Ã— 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 âˆ†OAM

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

(By Pythagoras Axiom)

(5)^{2} = (4)^{2} + OM^{2}

AM = Â½ AB

25 = 16 + OM^{2}

OM^{2} = 25 â€“ 16 = 9= (3)^{2}

OM = 3 cm

Again in right âˆ†OCN,

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

(5)^{2 }= (3)^{2 }+ ON^{2}

(CN = Â½ CD)

25 = 9 + ON^{2}

ON^{2} = 25 â€“ 9 = 16 = (4)^{2}

ON = 4

In fig (i), distance MN = ON â€“ OM

= 4 â€“ 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 âŠ¥ AB

M is midpoint of AB

AM = 12 cm

(i) Radius of circle OA = âˆšOM^{2 }+ AM^{2}

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

13^{2 }= 12^{2} + CN^{2}

CN = âˆš13^{2} â€“ 12^{2} = âˆš169 â€“ 144 = âˆš25

CN = 5 cm

As ON âŠ¥ CD, N is mid-Point of CD

CD = 2CN = 2 Ã— 5 = 10 cm

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

Let radius OB = OC = r

OE = (r-3) Cm.

Now in right âˆ†OBE

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

r^{2 }= (4)^{2} + (r â€“ 3)^{2}

r^{2 }= 16 + r^{2 }â€“ 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 âˆ¥ 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 âŠ¥ AB

AM = MB = 24/2 = 12 cm

Similarly ON âŠ¥ CD

CN = ND = 18/2 = 9 cm

Similarly In right âˆ† CNO

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

225 = 81 + ON^{2}

ON^{2 }= 225 â€“ 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 âŠ¥ CD which intersects AB in L.

Let OL =x, then OM = x + 3

Now right âˆ†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 âˆ†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 â€“ 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 = âˆš29cm

Diameter of the circle = 2 r

= 2 Ã— âˆš29 cm = 2 âˆš29 cm

**10. ABC is an isosceles triangle inscribed in a circle. If AB = AC = 12âˆš5 cm and BC = 24 cm, find the radius of the circle.**

**Solution : **

AB = AC 12âˆš5 and BC = 24 cm.

Join OB and OC and OA

Draw AD âŠ¥ BC which will pass through

Centre O

OD bisect BC in D

BD = DC = 12 cm

In right âˆ†ABD

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

(12âˆš5)^{2 } = AD^{2 }+ BD^{2}

(12âˆš5)^{2 }= AD^{2} + (12)^{2}

144 Ã— 5 =AD^{2} + 144

720 â€“ 144 = AD^{2}

AD^{2 }= 576 (AD = âˆš576 = 24

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

OD = AD â€“ AO = 24 â€“ r

Now in right âˆ†OBD,

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

r^{2 }= (12)^{2 } + (24 â€“ 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 âŠ¥ BC which will pass

Through the centre O of the circle.

Each side of âˆ†ABC = 6 cm.

AD = âˆš3/2 a= âˆš3/2 Ã— 6 = 3 âˆš3 cm.

OD = AD â€“ AO = 3âˆš3 â€“ r

Now in right âˆ†OBD

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

r^{2 }= (3)^{2 } + (3âˆš3-r)^{2}

r^{2 = }9 + 27 + r^{2 } -6 âˆš3r

(D is mid-point of BC)

6âˆš3r = 36

R = 36/6âˆš3 = 6/âˆš3 Ã— âˆš3/âˆš3 = 6âˆš3/3 = 2âˆš3 cm

Radius = 2âˆš3 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 âŠ¥ AB

Join OC

OM = AM â€“ AO = 18 â€“ 13 = 5 cm

OC = OA = 13 cm

Now in right âˆ†OMC

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

(13)^{2} = (5)^{2 } + MC^{2 }(MC^{2 }= 13^{2 }â€“ 5^{2}

MC^{2 }= 169 â€“ 25 = 144 = (12)^{2}

MC = 12

M is Mid-Point of CD

CD = 2 Ã— MC = 2 Ã— 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 âˆ 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

âˆ AOC = âˆ BOC

To prove : arc AC = arc BC

Proof : OC is the bisector of âˆ AOB

Or âˆ AOC = âˆ 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 âˆ AOB

I,e âˆ AOC = âˆ BOC

Proof : C is the mid-point of arc AB.

arc AC = arc BC

But arc AC and arc BC subtend âˆ AOC and

âˆ BOC at the centre

âˆ AOC = âˆ BOC

Hence OC Bisects the âˆ 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 â€“ 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 :**Â **(i) PQ**Â **(ii) AP**Â **(iii) BP**

**Solution:**

Given, radius = 15 cm

OA = OB = OP = OQ = 15 cm

Also, OM = 9 cm

MB = OB â€“ OM = 15 â€“ 9 = 6 cm

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

In âˆ†OMP, By using Pythagoras Theorem,

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

15^{2} = 9^{2 }+ PM^{2}

PM^{2} = 255 â€“ 81

PM = âˆš144 = 12 cm

Also, In âˆ†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 â€“ 81)

QM = âˆš144 = 12 cm

PQ = PM + QM

(As radius is bisected at M)

PQ = 12 + 12 cm = 24 cm

(ii) Now in âˆ†APM

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

AP^{2 }=24^{2} + 12^{2}

AP^{2 }= 576 + 144

AP = âˆš720 = 12 âˆš5 cm

(iii) Now In âˆ†BMP

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

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

BP^{2 }= 36 + 144

BP = âˆš180 = 6 âˆš5 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 âŠ¥ PS

QM = Â½ QR = Â½ Ã— 12 = 6 cm

In right âˆ†OQM

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

(10)^{2 } = OM^{2 } + (6)^{2}

OM^{2 } = 10^{2} â€“ 6^{2}

= 100 â€“ 36 = 64 = (8)^{2}

OM = 8 cm

Now in right âˆ†OPM

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

(17)^{2 } = OM^{2 }+ PM^{2}

PM^{2 }= (17)^{2} â€“ (8)^{2}

= 289 â€“ 64 = 225 = (15)^{2}

PM = 15 cm

PQ = PM â€“ QM = 15 â€“ 6 = 9 cm