In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 10 Circles Ex 10.2. These solutions are based on new NCERT Syllabus.

## NCERT Class 10th Maths Solutions Chapter 10 Circles Ex 10.2

In questions 1 to 3, choose the correct option and give justification.

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

(B) 12 cm

(C) 15 cm

(D) 24.5 cm

Solution:

(A): ∵ QT is a tangent to the circle at T and OT is radius

∴ OT⊥QT

Also, OQ = 25 cm and QT = 24 cm

∴ Using Pythagoras theorem, we get

OQ^{2} = QT^{2} + OT^{2}

⇒ OT^{2} = OQ^{2} – QT^{2} = 25^{2} – 24^{2} = 49

⇒ OT = 7

Thus, the required radius is 7 cm.

Question 2.

In figure, if TP and TQ are the two tangents to a circle with centre 0 so that ∠POQ =110°, then ∠PTQ is equal to

(A) 60°

(B) 70°

(C) 80°

(D) 90°

Solution:

(B): TQ and TP are tangents to a circle with centre O and ∠POQ = 110°

∴ OP⊥PT and OQ⊥QT

⇒ ∠OPT = 90° and ∠OQT = 90°

Now, in the quadrilateral TPOQ, we get

∠PTQ + 90° + 110° + 90° = 360° [Angle sum property of a quadrilateral]

⇒ ∠PTQ + 290° = 360°

⇒ ∠PTQ = 360° – 290° = 70°

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:

(A) : Since, O is the centre of the circle and two tangents from P to the circle are PA and PB.

∴ OA⊥AP and OB⊥BP

⇒ ∠OAP = ∠OBP = 90°

Now, in quadrilateral PAOB, we have

∠BPA + ∠PAO + ∠AOB + ∠OBP = 360°

⇒ 80° + 90° + ∠AOB + 90° = 360°

⇒ 260° + ∠AOB = 360°

⇒ ∠AOB = 360° – 260° ⇒ ∠AOB = 100°

In right ∆OAP and right ∆OBP, we have

OP = OP [Common]

∠OAP = ∠OBP [Each 90°]

OA = OB [Radii of the same circle]

∴ ∆OAP ≅ ∆OBP [By RHS congruency]

⇒ ∠POA = ∠POB [By CPCT]

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

Question 4.

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

Solution:

In the figure, PQ is diameter of the given circle and O is its centre.

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

Since, the tangents at a point to a circle is perpendicular to the radius through the point.

∴ PQ⊥AB

⇒ ∠APQ = 90°

And PQ⊥CD

⇒ ∠PQD = 90° ⇒ ∠APQ = ∠PQD

But they form a pair of alternate angles.

∴ AB || CD

Hence, the two tangents are parallel.

Question 5.

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

Solution:

In the figure, the centre of the circle is O and tangent AB touches the circle at P. If possible, let PQ be perpendicular to AB such that it is not passing through O.

Join OP.

Since, tangent at a point to a circle is perpendicular to the radius through that point.

∴ OP⊥AB

⇒ ∠OPB = 90° ……….. (1)

But by construction, PQ⊥AB

⇒ ∠QPB = 90° ………….. (2)

From (1) and (2),

∠QPB = ∠OPB

which is possible only when O and Q coincide. Thus, the perpendicular at the point of contact to the tangent to a circle passes through the centre.

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:

∵ The tangent to a circle is perpendicular to the radius through the point of contact.

∴ ∠OTA = 90°

Now, in the right ∆OTA, we have

OA^{2} = OT^{2} + AT^{2} [Pythagoras theorem]

⇒ OT^{2} = 5^{2} – 4^{2}

⇒ OT^{2} = (5 – 4)(5 + 4)

⇒ OT^{2} = 1 × 9 = 9 = 3^{2}

⇒ OT = 3

Thus, the radius of the circle is 3 cm.

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:

In the figure, O is the common centre, of the given concentric circles.

AB is a chord of the bigger circle such that it is a tangent to the smaller circle at P.

Since, OP is the radius of the smaller circle.

∴ OP⊥AB ⇒ ∠APO = 90°

Also, radius perpendicular to a chord bisects the chord.

∴ OP bisects AB

⇒ AP = \(\frac{1}{2}\) AB

Now, in right ∆APO,

OA^{2} = AP^{2} + OP^{2}

⇒ 5^{2} = AP^{2} + 3^{2} ⇒ AP^{2} = 5^{2} – 3^{2}

⇒ AP^{2} = 4^{2} ⇒ AP = 4 cm

⇒ \(\frac{1}{2}\) AB = 4 ⇒ AB = 2 × 4 = 8 cm

Hence, the required length of the chord AB is 8 cm.

Question 8.

A quadrilateral ABCD is drawn to circumscribe a circle (see figure).

Prove that AB + CD = AD + BC

Solution:

Since, the sides of quadrilateral ABCD, i.e., AB, BC, CD and DA touches the circle at P, Q, R and S respectively, and the lengths of two tangents to a circle from an external point are equal.

∴ AP = AS, BP = BQ,

DR = DS and CR = CQ

Adding them, we get

(AP + BP) + (CR + RD) = (BQ + QQ) + (DS + SA)

⇒ AB + CD = BC + DA

Question 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°.

Solution:

∵ The tangents drawn to a circle from an external point are equal.

∴ AP = AC ……… (1)

Join OC.

In ∆PAO and ∆CAO, we have

AO = AO [Common]

OP = OC [Radii of the same circle]

AP = AC [From (1)]

⇒ ∆PAO ≅ ∆CAO [SSS congruency]

∴ ∠PAO = ∠CAO

⇒ ∠PAC = 2∠CAO …………. (2)

Similarly, ∠CBQ = 2∠CBO ……………… (3)

Again, we know that sum of internal angles on the same side of a transversal is 180°.

∴ ∠PAC + ∠CBQ = 180°

2∠CAO + 2∠CBO = 180° [From (2) and (3)]

⇒ ∠CAO + ∠CBO = \(\frac{180^{\circ}}{2}\) = 90° …………… (4)

Also, in ∆AOB,

∠BAO + ∠OBA + ∠AOB = 180° [Sum of angles of a triangle]

⇒ ∠CAO + ∠CBO + ∠AOB = 180°

⇒ 90° + ∠AOB = 180° [From (4)]

⇒ ∠AOB = 180° – 90°

⇒ ∠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:

Let PA and PB be two tangents drawn from an external point P to a circle with centre O.

Now, in right ∆OAP and right ∆OBP, we have

PA = PB [Tangents to circle from an external point]

OA = OB [Radii of the same circle]

OP = OP [Common]

⇒ ∆OAP ≅ ∆OBP [By SSS congruency]

∴ ∠OPA = ∠OPB [By CPCT]

and ∠AOP = ∠BOP

⇒ ∠APB = 2∠OPA and ∠AOB = 2∠AOP

In right ∆OAP,

∠AOP + ∠OPA + ∠PAO = 180°

⇒ ∠AOP = 180° – 90° – ∠OPA

⇒ ∠AOP = 90° – ∠OPA

⇒ 2∠AOP = 180° – 2∠OPA

⇒ ∠AOB = 180° – ∠APB

⇒ ∠AOB + ∠APB = 180°

Question 11.

Prove that the parallelogram circumscribing a circle is a rhombus.

Solution:

We have ABCD, a parallelogram which circumscribes a circle (i.e., its sides touch the circle) with centre O.

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

∴ AP = AS

BP = BQ

CR = CQ

DR = DS

On adding, we get

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

⇒ AB + CD = AD + BC

But AB = CD [Opposite sides of parallelogram]

and BC = AD

∴ AB + CD = AD + BC ⇒ 2AB = 2BC

⇒ AB = BC

Similarly, AB = DA and DA = CD

Thus, AB = BC = CD = DA

Hence, 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:

Here ∆ABC circumscribes the circle with centre O. Also, radius = 4 cm

Let AC and AB touches the circle at E and F, respectively and join OE and OF.

∵ The sides BC, CA and AB touches the circle at D, E and F respectively.

∴ BF = BD = 8 cm

[ ∵ Tangents to a circle from an external point are equal]

CD = CE = 6 cm

AF = AE = x cm (say)

∴ The sides of the ∆ABC are 14 cm, (x + 6) cm and (x + 8) cm

Perimeter of ∆ABC

= [14 + (x + 6) + (x + 8)] cm

= [14 + 6 + 8 + 2x] cm

= (28 + 2x) cm

⇒ Semi perimeter of ∆ABC,

s = \(\frac{1}{2}\) [28 + 2x] cm = (14 + x) cm

∴ s – a = (14 + x) – (8 + x) = 6

s – b = (14 + x) – (14) = x

s – c = (14 + x) – (6 + x) = 8

where, a = AB, b = BC, c = AC

Squaring both sides, we get

(14 + x)^{2} = (14 + x)3x

⇒ 196 + x^{2} + 28x = 42x + 3x^{2}

⇒ 2x^{2} + 14x – 196 = 0

⇒ x^{2} + 7x – 98 = 0

⇒ (x – 7)(x + 14) = 0

⇒ x – 7 = 0 or x + 14 = 0

⇒ x = 7 or x = -14

But x = -14 is rejected.

∴ x = 7

Thus, AB = 8 + 7 = 15 cm, BC = 8 + 6 = 14 cm and CA = 6 + 7 = 13 cm

Question 13.

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

Solution:

We have a circle with centre O. A quadrilateral ABCD is such that the sides AB, BC, CD and DA touches the circle at P, Q, R and S respectively.

Join OP, OQ, OR and OS.

We know that 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

Also, the sum of all the angles around a point is 360°.

∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°

∴ 2(∠1 + ∠8 + ∠5 + ∠4) = 360°

⇒ (∠1 + ∠8 + ∠5 + ∠4) = 180° …………. (1)

and 2(∠2 + ∠3 + ∠6 + ∠7) = 360°

⇒ (∠2 + ∠3 + ∠6 + ∠7) = 180° ……………. (2)

Since, ∠2 + ∠3 = ∠AOB, ∠6 + ∠7 = ∠COD, ∠1 + ∠8 = ∠AOD and ∠4 + ∠5 = ∠BOC

∴ From (1) and (2), we have

∠AOD + ∠BOC = 180°

and ∠AOB + ∠COD = 180°