# NCERT solution for class 10 maths circles ( Chapter 10)

#### Solution for Exercise - 10.1

Q1. How many tangents can a circle have ?

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 ______.

(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

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

#### Solution for 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

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

∴ $$OT ⊥ QT$$

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

∴ By Pythagoras theorem, we have

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

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

∴ radius of the circle is 7 cm

∴ 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) $$90°$$

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

∴ $$OP \ ⊥ \ PT$$ and $$OQ \ ⊥ \ QT$$

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

In the quadrilateral TPOQ, we have

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

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

$$∠ \ PTQ \ + \ 290° \ = \ 360°$$

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

$$=> \ ∠ \ PTQ \ = \ 70°$$

∴ 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) $$80°$$

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

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

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

In the quadrilateral PAOB, we have

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

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

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

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

In $$∆ \ OAP$$ and $$OBP$$, we have

$$OP \ = \ OP$$      [Common]

$$OA \ = \ OB$$      [Radii]

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

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

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

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

∴ The (A) is the correct option.

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

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.

∵ Tangent at a point to a circle is perpendicular to the radius through the point,

∴ $$PQ \ ⊥ \ AB$$ and $$PQ \ ⊥ \ CD$$

$$=> \ ∠ \ APQ \ = \ ∠ \ PQD$$

$$=> \ AB \ || \ CD$$      [∵ $$∠ \ APQ$$ 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.

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.

∵ Tangent at a point to a circle is perpendicular to the radius through the point,

∴ $$AB \ ⊥ \ OP$$ => $$∠ \ OPB \ = \ 90°$$

$$∠ \ QPB \ = \ 90°$$      (Construction)

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

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.

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

∴ $$∠ \ OTA \ = \ 90°$$

In $$∆ \ OTA$$, we have

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

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

∴ 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.

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.

∵ 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

Q.9 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 $$∠ \ AOB \ = \ 90°$$.

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

∴ $$AP \ = \ AC$$.
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]

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

Similarly, we can prove that

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

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

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

=> $$∠ \ CAO \ + \ ∠ \ CBO \ = \ 90°$$
=> $$180° \ - \ ∠ \ AOB \ = \ 90°$$      [∵ $$∠ \ CAO \ + \ ∠ \ CBO \ + \ ∠ \ AOB \ = \ 180°$$]

$$=> \ ∠ \ AOB \ = \ 90°$$

Q.10 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 centre.

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]

$$OA \ = \ OB$$     [Radii of the same circle]

$$OP \ = \ OP$$     [Common]

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

=> $$∠ \ OPA \ = \ ∠ \ OPB$$

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

and, $$∠ \ AOB \ = \ 2 ∠ \ AOP$$

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

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

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

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

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.

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 $$CR \ = \ CQ$$

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

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

$$=> \ 2AB \ = \ 2BC$$      [∵ ABCD is a parallelogram => $$AB \ = \ CD$$ and $$BC \ = \ AD$$]

=> $$AB \ = \ BC$$

∴ $$AB \ = \ BC \ = \ CD \ = \ AD$$

∴ ABCD is a rhombus.

Q.12 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.

Let us join AO, OC, and OB.

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

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

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

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

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

$$=> \ s \ = \ 14 \ + \ x$$

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

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

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

∴ 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 $$∆ \ ABC \ = \ Area (∆ OBC) \ + \ Area (∆ OCA) \ + \ Area (∆ OAB)$$

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

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

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

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

Squaring, we get

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

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

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

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

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

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

But x cannot be negative,

∴$$x \ = \ 7$$

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.

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.

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

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

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

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

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

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

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.