These NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 10 Maths Chapter 10 Circles Exercise 10.2

Question 1.

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 Sol.

(b) 12 cm

(c) 15 cm

(d) 24.5 cm

Solution:

Therefore, radius of the circle is 7 cm

So, option (A) of the give question is right.

Question 2.

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°

Solution:

∠OPT = 90°

∠OQT = 90°

∠POQ = 110°

TPOQ is a quadrilateral,

∴ ∠PTQ + ∠POQ = 180° ⇒ ∠PTQ + 110° = 180°

⇒ ∠PTQ = 180°- 110° = 70°

Hence, correct option is (b).

Question 3.

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°

Solution:

In AOAP and AOBP

OA = OB [Radii]

PA = PB

[Lengths of tangents from an external point are equal]

OP = OP [Common]

∴ ∆OAP ≅ ∆OBP [SSS congruence rule]

∠AOB + ∠APB = 180° ⇒ ∠AOB + 80° = 180°

⇒ ∠AOB = 180° – 80° = 100°

From eqn. (i), we get

⇒ ∠POA = \(\frac { 1 }{ 2 }\) x 100° = 50°

Hence, correct option is (a)

Question 4.

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

Solution:

AB is a diameter of the circle, p and q are two tangents.

OA ⊥ p and OB ⊥ q

∠1 = ∠2 = 90°

⇒ p || q ∠1 and ∠2 are alternate angles]

Question 5.

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

Solution:

XY tangent to the circle C(0, r) at B and AB ⊥ XY. Join OB.

∠ABY = 90° [Given]

∠OBY = 90°

[Radius through point of contact is perpendicular to the tangent]

∴ ∠ABY + ∠OBY = 180° ⇒ AB Oiscollinear

∴ AB passes through centre of the circle.

Question 6.

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.

Solution:

OA = 5 cm, AP = 4 cm OP = Radius of the circle

∠OPA = 90° [Radius and tangent are perpendicular]

Question 7.

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.

Solution:

Radius of larger circle = 5 cm Radius of smaller circle = 3 cm

OP ⊥ AB

[Radius of circle is perpendicular to the tangent]

AB is a chord of the larger circle

Question 8.

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

Solution:

AP = AS … (i)

[Lengths of tangents from an external point are equal]

BP = BQ … (ii)

CR = CQ … (iii)

DR = DS … (iv)

Adding equations (i), (ii), (iii) and (iv), we get

AP + BP + CR + DR = AS + BQ + CQ + DS

⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

⇒ AB + CD = AD + BC

Hence proved.

Question 9.

In figure, XY and X’Y’ are two parallel tangents to a circle , x 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°.

Solution:

To Prove : ∠AOC = 90°

Join OC

XY and X’Y’ are two tangents

XY || X’Y’ (given) and AB is transversal … (i)

∠PAC + ∠QBC = 180° (Sum of int ∠s)

In ∆PAO and ∆CAO, we have

PA = AC (Tangent drawn from exterior point)

AO = AO (common)

PO = QC (radii of same circle)

∆ ∆PAO ≅ ∆CDO

⇒ ∠1 = ∠2

Similarly ∠3 = ∠4

Now, ∠PAC + ∠QBC = 180°

⇒ ∠1 + ∠2 + ∠3 + ∠4 = 180°

[∵ ∠1 = ∠2 and ∠4 = ∠3]

⇒ 2(∠2 + ∠3) = 180°

or ∠2 + ∠3 = 90°

Now, In ∆ OAB,

∠AOB + ∠2 + ∠3= 180° [put ∠2 + ∠3 = 90°]

or ∠AOB+ 90°= 180°

⇒ ∠AOB = 90°

Question 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.

Solution:

Given : Two tangents PA and PB are drawn from the point P to the circle with centre O. A and B are the points of contact of two tangents PA and PB respectively.

To Prove : ∠APB + ∠AOB = 180°

Construction: Join OA and OB

Proof : Since the tangent PA is drawn to the circle at A with centre O, we have;

OA ⊥ PA ⇒ ∠OAP = 90°

Similarly, since the tangent PB is drawn to the circle at B with centre O. we have,

OB ⊥ PB ⇒ OBP = 90°

We know the sum of four angles of a quadrilateral is 360°

Sc, in the quadrilateral, OA PB, we have

∠OAP + ∠APB + ∠OBP + ∠AOB = 360°

⇒ ∠APB + ∠AOB = 360° (∠OAP + ∠OBP) = 360° – 180°

⇒ ∠APB + ∠AOB = 180° Hence Proved.

Question 11.

Prove that the parallelogram circumscribing a circle is a rhombus.

Solution:

Given: ABCD is a || gm AB, BC, CD and DA are touching a circle.

To Prove: AB = BC = CD = DA

Proof : Since the length of two tangents drawn from an external point to a circle are equal, we have

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

AB + CD = AP + BP + CR + DR

AD + BC = AS + BQ + CQ + DS

= AS + DS + BQ + CQ

= AD + BC

⇒ AB + CD = AD + BC

⇒ 2AB = 2AD

w AB = AD

But AB= CD and BC = AD

AB = BC = CD = DA

Hence || gm ABCD is a rhombus.

Question 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.

Solution:

Let the sides AB, BC and AC touch the in circle at E, D and F respectievly. Join the centre O of the circle with A, B, C, D, E and F.

The point D divide BC in two parts CD = 6 cm and BD = 8 cm

Since tangent to a circle from an external point are equal CD = CF = 6 cm and DB = BE = 8 cm.

AF = AE = x cm (say)

OD= OE = OF = 4 [radii of the incircle]

Now, Area of ∆AOC = \(\frac { 1 }{ 2 }\) (6 + x) x 4

= (6 + x) × 2 = (12 + 2x) sq. cm. … (i)

Similarly Area of ∆AOB = \(\frac { 1 }{ 2 }\) (8 + x) × 4

= (16 + 2x) sq. cm … (ii)

Area of ABOC = \(\frac { 1 }{ 2 }\) (6 + 8) x 4 = 28 sq. cm … (iii)

Adding equations (i), (ii) and (iii), we have

Area of ∆ABC = [(12 + 2x) + (16 + 2x) + 28] sq.

= [56 + 4x] sq. cm … (iv)

Again perimeter of ∆ABC = AB + BC + CA = (x + 8) + (6 + 8) + (6 + x)

= (28 + 2x) cm

AB = x + 8 = 7 + 8 = 15 cm

AC = x + 6 = 7 + 6 = 13 cm.

Question 13.

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

Solution:

Given : A circle with centre O touches the side, PQ, QR, RS and PS of quadrilateral at the points A, B, C and D respectively.

To Prove : ∠PQQ + ∠ROS = 180°

and ∠POS + ∠QOR = 180°

Construction: Join OA, OB, OC and OD.

Proof: In ∆POD and ∆POA .

PA = FD [tangents from expemal point]

OP = OP [common]

∠ODP = ∠ODA [each 90°]

∆POD = ∆POA [RHS congruency]

∠1 = ∠2 [C.P.C.T]

Similarly we have

∠3= ∠4; ∠5 = ∠6 and ∠7 = ∠8

Now ∠1 + ∠2 + ∠3+ ∠4 + ∠5 +

∠6 + ∠7 + ∠8 = 360°

Using the above result

∠2 + ∠2 + ∠3+ ∠3+ ∠6 + ∠6 + ∠7 + ∠7 = 360°

or 2(∠2 + ∠3)+ (∠6 + ∠7)

\(\frac{360^{\circ}}{2}\) = 180°

and

∠1 + ∠1 + ∠4 + ∠4 + ∠5 + ∠5 + ∠8 + ∠8 = 360°

or 2(∠1 + ∠4 + ∠5 + ∠8) = 360°

or ∠1 + ∠4 + ∠5 + ∠8 = \(\frac{360^{\circ}}{2}\)

or (∠1 + ∠8) + (∠4 + ∠5) = 180°

or ∠POS +∠QOR = 180°