# NCERT Solutions for Class 10 Maths Chapter 10 Circles

Written by Team Trustudies
Updated at 2021-05-07

## NCERT solutions for class 10 Maths Chapter 10 Circles Exercise - 10.1

Q1 ) How many tangents can a circle have ?

NCERT Solutions for Class 10 Maths Chapter 10 Circles

A circle can have an infinite number of tangents.

Q2 ) Fill in the blanks :
(i) A tangent to a circle intersects it in _________ point(s).

(ii) A line intersecting a circle in two points is called a _______.

(iii) A circle can have _______ parallel tangents at the most.

(iv) The common point of a tangent to a circle and the circle is called ______.

NCERT Solutions for Class 10 Maths Chapter 10 Circles

(i) exactly one

(ii) secant

(iii) two

(iv) point of contact

Q3 ) A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length of PQ is :
(A) $$12$$ cm

(B) $$13$$ cm

(C) $$8.5$$ cm

(D) $$\sqrt{119}$$ cm

NCERT Solutions for Class 10 Maths Chapter 10 Circles

$$PQ \ = \ \sqrt{OQ^2 \ - \ OP^2} \$$
$$= \ \sqrt{12^2 \ - \ 5^2} \$$
$$= \ \sqrt{144 \ - \ 25} \ \$$
$$= \ \sqrt{119}$$cm

So, (D) option is correct

Q4 ) Draw a circle and two lines parallel to a given line such that one is tangent and the other, a secant to the circle.

NCERT Solutions for Class 10 Maths Chapter 10 Circles

## NCERT solutions for class 10 Maths Chapter 10 Circles Exercise - 10.2

Q1 ) 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

NCERT Solutions for Class 10 Maths Chapter 10 Circles

Since QT is a tangent to the circle at T and OT is radius,

$$\therefore \ OT ⊥ QT$$

It is given that OQ = 25 cm and QT = 24 cm,

$$\therefore$$ By Pythagoras theorem, we have

$$\Rightarrow OQ^2 \ = \ QT^2 \ + \ OT^2$$
$$\Rightarrow OT^2 \ = \ OQ^2 \ - \ QT^2$$
$$\Rightarrow \ OT^2 \ = \ 25^2 \ - \ 24^2$$
$$\Rightarrow OT^2 \ = \ 49$$

$$\Rightarrow \ OT \ = \ \sqrt{49} \ = \ 7$$

$$\therefore$$ radius of the circle is 7 cm

$$\therefore$$ Option (A) is correct.

Q2 ) In figure, if TP and TQ are the two tangents to a circle with centre O so that $$∠ \ POQ \ = \ 110°$$ , then $$∠ \ PTQ$$ is equal to
(A) $$60°$$
(B) $$70°$$
(C) $$80°$$
(D) $$90°$$

NCERT Solutions for Class 10 Maths Chapter 10 Circles

Since TP and TQ are tangents to a circle with centre O so that $$\angle \ POQ \ = \ 110°$$,

$$\therefore$$$$OP \ ⊥ \ PT$$ and $$OQ \ ⊥ \ QT$$

$$\Rightarrow \ ∠ \ OPT \ = \ 90°$$ and $$∠ \ OQT \ = \ 90°$$

In the quadrilateral TPOQ, we have

$$∠ \ PTQ \ + \ ∠ \ TPO \ + \ ∠ \ POQ \ + \ ∠ \ OQT \ = \ 360°$$
[ \therefore Sum of all angles in a quadrilateral = 360° ]

$$∠ \ PTQ \ + \ 90° \ + \ 110° \ + \ 90° \ = \ 360°$$

$$∠ \ PTQ \ + \ 290° \ = \ 360°$$

$$\Rightarrow \ ∠ \ PTQ \ = \ 360° \ - \ 290°$$

$$\Rightarrow \ ∠ \ PTQ \ = \ 70°$$

$$\therefore$$ Option (B) is correct.

Q3 ) If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then $$∠ \ POA$$ is equal to
(A) $$50°$$
(B) $$60°$$
(C) $$70°$$
(D) $$80°$$

NCERT Solutions for Class 10 Maths Chapter 10 Circles

Since PA and PB are tangents to a circle with centre O,

$$\therefore OA \ ⊥ \ AP$$, and $$OB \ ⊥ \ BP$$

$$∠ \ OAP \ = \ 90°$$ and $$∠ \ OBP=90°$$

In the quadrilateral PAOB, we have

$$\Rightarrow ∠ \ ABP \ + \ ∠ \ PAO \ + \ ∠ \ AOB \ + \ ∠ \ OBP \ = \ 360°$$

$$\Rightarrow \ ∠ \ 80° \ + \ 90° \ + \ ∠ \ AOB \ + \ 90° \ = \ 360°$$

$$\Rightarrow \ 260° \ + \ ∠ \ AOB \ = \ 360°$$

$$\Rightarrow \ ∠ \ AOB \ = \ 360° \ - \ 260° \ = \ 100°$$

In $$∆ \ OAP$$ and $$OBP$$, we have

$$OP \ = \ OP$$ [Common]

$$OA \ = \ OB$$ [Radii]

$$∠ \ OAP \ = \ ∠ \ OBP$$ [Each = 90°]

$$\therefore$$ $$∆ \ OAP \ \cong \ ∆ \ OBP$$ ( By SAS Criterion)

$$\Rightarrow \ ∠ \ POA \ = \ ∠ \ POB$$ [C.P.C.T.]

$$\therefore$$ $$∠ \ POA \ = \ \frac{1}{2} \ ∠ \ AOB \$$
$$= \ \frac{1}{2} \ × \ 100° \ = \ 50°$$

∴ The (A) is the correct option.

Q4 ) Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

NCERT Solutions for Class 10 Maths Chapter 10 Circles

Let PQ be a diameter of the given circle with centre O.

Let AB and CD be the tangents drawn to the circle at the end points of the diameter PQ respectively.

$$\therefore$$ Tangent at a point to a circle is perpendicular to the radius through the point,

$$\therefore$$ $$PQ \ ⊥ \ AB$$ and $$PQ \ ⊥ \ CD$$

$$\Rightarrow \ ∠ \ APQ \ = \ ∠ \ PQD$$

$$\Rightarrow \ AB \ || \ CD$$ [$$\because$$ $$∠ \ APQ$$ and $$∠ \ PQD$$ are alternate angles]

Hence, the tangents drawn at the ends of a diameter of a circle are parallel.

Q5 ) Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

NCERT Solutions for Class 10 Maths Chapter 10 Circles

Let AB be the tangent drawn at the point P on the circle with O.

If possible, let PQ be perpendicular to AB, not passing through O. Join OP.

$$\therefore$$ Tangent at a point to a circle is perpendicular to the radius through the point,

$$\therefore$$ $$AB \ ⊥ \ OP$$
$$\Rightarrow ∠ \ OPB \ = \ 90°$$

$$∠ \ QPB \ = \ 90°$$ (Construction)

$$\therefore ∠ \ QPB \ = \ ∠ \ OPB$$ , which is not possible.

$$\therefore$$ It contradicts our supposition.

Hence, the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Q6 ) The Length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

NCERT Solutions for Class 10 Maths Chapter 10 Circles

Since tangent to a circle is perpendicular to the radius through the point of contact,

$$\therefore$$ $$∠ \ OTA \ = \ 90°$$

In $$∆ \ OTA$$, we have

$$\Rightarrow OA^2 \ = \ OT^2 \ + \ AT^2 \$$
$$\Rightarrow \ 5^2 \ = \ OT^2 \ + \ 4^2$$
$$\Rightarrow \ OT^2 \ = \ 5^2 \ - \ 4^2 \ = \ 25 \ - \ 16 \ = \ 9$$

$$OT \ = \ \sqrt{9} \ = \ 3$$

$$\therefore$$ Radius of the circle is 3 cm.

Q7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

NCERT Solutions for Class 10 Maths Chapter 10 Circles

Let O be the common centre of two concentric circles, and let AB be a chord of the larger circle touching the smaller circle at P. Join OP.

∵ OP is the radius of the smaller circle and AB is tangent to this circle at P,

∴ $$OP \ ⊥ \ AB$$.
We know that, the perpendicular drawn from the centre of a circle to any chord of the circle bisects the chord.

So, $$OP \ ⊥ \ AB$$ and $$AP \ = \ BP$$.

In $$∆ \ APO$$, we have

$$OA^2 \ = \ AP^2 \ + \ OP^2 \ => \ 5^2 \ = \ AP^2 \ + \ 3^2$$

$$AP^2 \ = \ 5^2 \ - \ 3^2 \ = \ 25 \ - \ 9 \ = \ 16$$
$$AP \ = \ 4$$

Now, $$AB \ = \ 2AP$$      [∵ $$AP \ = \ BP$$ ]

$$AB = \ 2 \ × \ 4 \ = \ 8$$

∴ The length of the chord of the larger circle which touches the smaller circle is 8 cm.

Q8 ) A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove that AB + CD = AD + BC.

NCERT Solutions for Class 10 Maths Chapter 10 Circles

∵ Lengths of two tangents drawn from an external point of circle are equal,

$$AP \ = \ AS$$ , $$BP \ = \ BQ$$ , $$DR \ = \ DS$$ , and $$CR \ = \ CQ$$

$$(AP \ + \ BP) \ + \ (CR \ + \ RD) \$$
$$= \ (BQ \ + \ QC) \ + \ (DS \ + \ SA)$$

$$=> \ AB \ + \ CD \ = \ BC \ + \ DA$$

Hence proved

Q9 ) In figure, XY and X'Y' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X'Y' at B. Prove that $$∠ \ AOB \ = \ 90°$$.

NCERT Solutions for Class 10 Maths Chapter 10 Circles

Since tangents drawn from an external point to a circle are equal,

$$\because$$ $$AP \ = \ AC$$.
Thus in $$∆ \ PAO$$ and $$∠ \ AOC$$, we have

$$AP \ = \ AC$$
$$AO \ = \ AO$$ [Common]
$$PO \ = \ OC$$ [Radii of the same circle]
$$∆ \ PAO \ \cong \ AOC$$ [ By SSS-criterion of congruence]

$$∠ \ PAO \ = \ ∠ \ CAO \$$
$$\Rightarrow \ ∠ \ PAC \ = \ 2 ∠ CAO$$

Similarly, we can prove that

$$∠ \ QBO \ = \ ∠ \ CBO \$$
$$\Rightarrow \ ∠ \ CBQ \ = \ 2 ∠ \ CBO$$

Now, $$∠ \ PAC \ + \ ∠ \ CBQ \ = \ 180°$$
[Sum of the interior angles on the same side of transversal is 180°]

$$\Rightarrow 2 ∠ \ CAO \ + \ 2 ∠ \ CBO \ = \ 180°$$

$$\Rightarrow ∠ \ CAO \ + \ ∠ \ CBO \ = \ 90°$$
$$\Rightarrow 180° \ - \ ∠ \ AOB \ = \ 90°$$
[$$\because$$ $$∠ \ CAO \ + \ ∠ \ CBO \ + \ ∠ \ AOB \ = \ 180°$$]

$$\Rightarrow \ ∠ \ AOB \ = \ 90°$$

Q10 ) Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the center.

NCERT Solutions for Class 10 Maths Chapter 10 Circles

Let PA and PB be two tangents drawn from an external point P to a circle with centre O. We have to prove that- $$∠ \ AOB \ + \ ∠ \ APB \ = \ 180°$$

In right $$∆ \ OAP$$ and $$OBP$$, we have

$$PA \ = \ PB$$
[Tangents drawn from an external point are equal]

$$OA \ = \ OB$$

$$OP \ = \ OP$$ [Common]

$$\therefore ∆ \ OAP \ \cong \ ∆ \ OBP$$
[by SSS - criterion of congruence]

$$\Rightarrow ∠ \ OPA \ = \ ∠ \ OPB$$

and, $$∠ \ AOP \ = \ ∠ \ BOP \$$
$$\Rightarrow \ ∠ \ APB \ = \ 2 ∠ \ OPA$$

and, $$∠ \ AOB \ = \ 2 ∠ \ AOP$$

But, $$∠ \ AOP \ = \ 90° \ - \ ∠ \ OPA$$
[ $$\because ∆ \ OAP$$ is right triangle]

$$\therefore 2 ∠ \ AOP \ = \ 180° \ - \ 2 ∠ \ OPA$$

$$\Rightarrow ∠ \ AOB \ = \ 180° \ - \ ∠ \ APB$$

$$\Rightarrow ∠ \ AOB \ + \ ∠ \ APB \ = \ 180°$$

Hence proved that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.

Q11 ) Prove that the parallelogram circumscribing a circle is a rhombus.

NCERT Solutions for Class 10 Maths Chapter 10 Circles

Let ABCD be a parallelogram such that its sides touch a circle with centre O.

We know that the tangents to a circle from an exterior point are equal in length.

$$AP \ = \ AS$$ , $$BP \ = \ BQ$$ , $$DR \ = \ DS$$ , and $$CR \ = \ CQ$$

$$(AP \ + \ BP) \ + \ (CR \ + \ RD) \ = \ (BQ \ + \ QC) \ + \ (DS \ + \ SA)$$

$$\Rightarrow \ AB \ + \ CD \ = \ BC \ + \ DA$$

$$\Rightarrow \ 2AB \ = \ 2BC$$ [$$\because$$ ABCD is a parallelogram ]
$$\Rightarrow AB \ = \ CD$$ and $$BC \ = \ AD$$]

$$\Rightarrow AB \ = \ BC$$

$$\therefore$$ $$AB \ = \ BC \ = \ CD \ = \ AD$$

$$\therefore$$ ABCD is a rhombus.

Q12 ) A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC.

NCERT Solutions for Class 10 Maths Chapter 10 Circles

Let us join AO, OC, and OB.

It is given that BD = 8cm, CD = 6 cm.

$$\therefore$$ Lengths of two tangents drawn from an external point of circle are equal.

$$\therefore$$ $$BF \ = \ BD \ = \ 8 cm \ ,$$
$$\ CE \ = \ CD \ = \ 6 cm \ ,$$
$$\ and \ let \ AF \ = \ AE \ = \ x$$ cm.

Then, the sides of the triangle are 14 cm, (x+6) cm and (x+8) cm.

$$\therefore 2s \ = \ 14 \ + \ (x + 6) \ + \ (x+8) \ = \ 28 \ + \ 2x$$

$$\Rightarrow \ s \ = \ 14 \ + \ x$$

$$s \ - \ a \ = \ 14 \ + \ x \ - \ 14 \ = \ x$$,

$$s \ - \ b \ = \ 14 \ + \ x \ - \ x \ - \ 6 \ = \ 8$$

and, $$s\ - \ c \ = \ 14 \ + \ x \ - \ x \ - \ 8 \ = \ 6$$

$$\therefore$$ Area of $$∆ \ ABC \$$
$$= \ \sqrt{s(s \ - \ a)(s \ - \ b)(s \ - \ c)}$$
$$= \ \sqrt{(14 \ + \ x)(x)(8)(6)}$$

$$= \ \sqrt{48(x^2 \ + \ 14x)}$$

Also, Area of $$∆ \ ABC \$$
$$= \ Area (∆ OBC) \ + \ Area (∆ OCA) \ + \ Area (∆ OAB)$$

$$= \ \frac{1}{2} \ × \ BC \ × \ OD \ + \ \frac{1}{2} \ × \ CA \ × \ OE \ + \ \frac{1}{2} \ × \ AB \ × \ OF$$

$$= \ \frac{1}{2} \ × \ 14 \ × \ 4 \ + \ \frac{1}{2} \ × \ (x+6) \ × \ 4 \ + \ \frac{1}{2} \ × \ (x+8) \ × \ 4$$

$$= \ 2(14 \ + \ x \ + \ 6 \ + \ x \ + \ 8) \ = \ 2(28 \ + \ 2x)$$

$$\therefore \sqrt{(48(x^2 \ + \ 14x)} \ = \ 4(14 \ + \ x)$$

Squaring, we get

$$48(x^2 \ + \ 14x) \ = \ 16(14 \ + \ x)^2$$

$$\Rightarrow \ 3(x^2 \ + \ 14x) \ = \ 196 \ + \ 28x \ + \ x^2$$

$$\Rightarrow 2x^2 \ + \ 14x \ - \ 196 \ = \ 0$$

$$\Rightarrow \ x^2 \ + \ 7x \ - \ 98 \ = \ 0$$

$$\Rightarrow \ (x \ - \ 7)(x \ + \ 14) \ = \ 0$$

$$\Rightarrow \ x \ = \ 7 \ or \ x \ = \ -14$$

But x cannot be negative,

$$\therefore x \ = \ 7$$

Thus, AB = x + 8 = 7 + 8 = 15 cm and AC = x + 6 = 6 + 7 = 13 cm.

Hence, the sides AB and AC are 15 cm and 13 cm respectively.

Q13 ) Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

NCERT Solutions for Class 10 Maths Chapter 10 Circles

Let a circle with centre O touch the sides AB, BC, CD and DA of a quadrilateral ABCD at the points P, Q, R and S respectively.

We have to prove that-

$$∠ \ AOB \ + \ ∠ \ COD = \ 180°$$

and, $$∠ \ AOD \ + \ ∠ \ BOC \ = \ 180°$$
Join OP, OQ, OR and OS.

$$\therefore$$ The two tangents drawn from an external point to a circle subtend equal angles at the centre.

$$\therefore$$ $$∠ \ 1 \ = \ ∠ \ 2 \ , \ ∠ \ 3 \ = \ ∠ \ 4 \ , \ ∠ \ 5 \ = \ ∠ \ 6 \ and \ ∠ \ 7 \ = \ ∠ \ 8$$

Now, $$∠ \ 1 \ + \ ∠ \ 2 \ + \ ∠ \ 3 \ + \ ∠ \ 4 \ + \ ∠ \ 5 \ + \ ∠ \ 6 \ + \ ∠ \ 7 \ + \ ∠ \ 8 \ = \ 360°$$ [$$\because$$ Sum of all the ∠ s around a point is 360°]

$$\Rightarrow \ 2(∠ \ 2 \ + \ ∠ \ 3 \ + \ ∠ \ 6 \ + \ ∠ \ 7) \ = \ 360°$$

and, $$2(∠ \ 1 \ + \ ∠ \ 8 \ + \ ∠ \ 4 \ + \ ∠ \ 5) \ = \ 360°$$

$$\Rightarrow \ (∠ \ 2 \ + \ ∠ \ 3 \ + \ ∠ \ 6 \ + \ ∠ \ 7) \ = \ 180°$$

and $$(∠ \ 1 \ + \ ∠ \ 8) \ + \ (∠ \ 4 \ + \ ∠ \ 5) \ = \ 180°$$

Now since $$∠ 2 + ∠ 3 = ∠ AOB$$ ,
$$\ ∠ 6 + ∠ 7 = ∠ COD,$$
$$\ ∠ 1 + ∠ 8 = ∠ AOD \$$ and $$\ ∠ 4 + ∠ 5 = ∠ BOC,$$ we can say that

$$∠ \ AOB \ + \ ∠ \ COD \ = \ 180°$$
and, $$∠ \ AOD \ + \ ∠ \ BOC \ = \ 180°$$

Hence proved that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

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