NCERT solution for class 10 maths circles ( Chapter 10)

Solution for Exercise - 10.1

Q1. How many tangents can a circle have ?
Answer :

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 ______.
Answer :


(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
Answer :




\( PQ \ = \ \sqrt{OQ^2 \ - \ OP^2} \ = \ \sqrt{12^2 \ - \ 5^2} \ = \ \sqrt{144 \ - \ 25} \ \ = \ \sqrt{119} \)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.
Answer :



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
Answer :




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

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

\( => \ OT \ = \ \sqrt{49} \ = \ 7 \)

∴ 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) \( 90° \)
Answer :




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) \( 80° \)
Answer :




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.
Answer :




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.
Answer :




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.

∴ 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.
Answer :


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.
Answer :


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, \( AB \ = \ 2AP \)      [∵ \( AP \ = \ BP \)]

\( = \ 2 \ × \ 4 \ = \ 8 \)

∴ 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.

Answer :


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

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

Adding these all, we get

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

\( => \ AB \ + \ CD \ = \ BC \ + \ DA \)

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 \( ∠ \ AOB \ = \ 90° \).

Answer :


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°]

\( 2 ∠ \ CAO \ + \ 2 ∠ \ CBO \ = \ 180° \)

=> \( ∠ \ CAO \ + \ ∠ \ CBO \ = \ 90° \ => \ 180° \ - \ ∠ \ AOB \ = \ 90° \)      [∵ Sum of all angles of a triangle is 180° => \( ∠ \ CAO \ + \ ∠ \ CBO \ + \ ∠ \ AOB \ = \ 180° \)]

\( => \ ∠ \ AOB \ = \ 90° \)

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.
Answer :


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]

∴ \( ∆ \ AOP \ \cong \ ∆ \ OBP \)     [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]

∴ \( 2 ∠ \ AOP \ = \ 180° \ - \ 2 ∠ \ OPA \)

=> \( ∠ \ AOB \ = \ 180° \ - \ ∠ \ APB \)

=> \( ∠ \ AOB \ + \ ∠ \ APB \ = \ 180° \)

Q11. Prove that the paralelogram circumscribing a circle is a rhombus.
Answer :


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 \(CR \ = \ CQ \)

Adding these all, we get

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

\( => \ AB \ + \ CD \ = \ BC \ + \ DA \)

\( => \ 2AB \ = \ 2BC \)      [∵ ABCD is a \( ||^{gm} \) => \( AB \ = \ CD \) 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.

Answer :


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

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

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

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

∴ \( \sqrt{(48(x^2 \ + \ 14x)} \ = \ 2(28 \ + \ 2x) \)

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,

∴\( x \ = \ 7 \)

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.
Answer :


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

\(∠ \ AOB \ + \ ∠ \ COD \ = \ 180° \)

and, \(∠ \ AOD \ + \ ∠ \ BOC \ = \ 180° \)      [∵ \(∠ \ 2 \ + \ ∠ \ 3 \ = \ ∠ \ AOB \ , \ ∠ \ 6 \ + \ ∠ \ 7 \ = \ ∠ \ COD \ ∠ \ 1 \ + \ ∠ \ 8 \ = \ ∠ \ AOD \ and \ ∠ \ 4 \ + \ ∠ \ 5 \ = \ ∠ \ BOC \) ]