# RD Sharma Solutions for Class 10 Maths Chapter 10 Circles

### RD Sharma Class 10 Chapter 10 Exercise 10.1 Page No: 10.5

**1. Fill in the blanks:**

**(i) The common point of tangent and the circle is called _________. **

**(ii) A circle may have _____ parallel tangent. **

**(iii) A tangent to a circle intersects it in ______ point. **

**(iv) A line intersecting a circle in two points is called a _______ **

**(v) The angle between tangent at a point P on circle and radius through the point is _______ **

**Solution:**

(i) The common point of tangent and the circle is called point of contact.

(ii) A circle may have two parallel tangent.

(iii) A tangent to a circle intersects it in one point.

(iv) A line intersecting a circle in two points is called a secant.

(v) The angle between tangent at a point P on circle and radius through the point is 90° .

**2. How many tangents can a circle have?**

**Solution:**

A tangent is defined as a line intersecting the circle in one point. Since, there are infinite number of points on the circle, a circle can have many (infinite) tangents.

### RD Sharma Class 10 Chapter 10 Exercise 10.2 Page No: 10.33

**1. If PT is a tangent at T to a circle whose centre is O and OP = 17 cm, OT = 8 cm. Find the length of the tangent segment PT.**

**Solution:**

Given,

OT = radius = 8 cm

OP = 17 cm

To find: PT = length of tangent =?

Clearly, T is point of contact. And, we know that at point of contact tangent and radius are perpendicular.

∴ OTP is right angled triangle ∠OTP = 90°, from Pythagoras theorem OT2 + PT2 = OP2

82 + PT2 = 172

∴ PT = length of tangent = 15 cm.

**2. Find the length of a tangent drawn to a circle with radius 5cm, from a point 13 cm from the center of the circle.**

**Solution:**

Consider a circle with centre O.

OP = radius = 5 cm. (given)

A tangent is drawn at point P, such that line through O intersects it at Q.

And, OQ = 13cm (given).

To find: Length of tangent PQ =?

We know that tangent and radius are perpendicular to each other.

∆OPQ is right angled triangle with ∠OPQ = 90°

By Pythagoras theorem we have,

OQ2 = OP2 + PQ2

⇒ 132 = 52 + PQ2

⇒ PQ2 = 169 – 25 = 144

⇒ PQ = √114

= 12 cm

Therefore, the length of tangent = 12 cm

**3. A point P is 26 cm away from O of circle and the length PT of the tangent drawn from P to the circle is 10 cm. Find the radius of the circle.**

**Solution:**

Given, OP = 26 cm

PT = length of tangent = 10 cm

To find: radius = OT = ?

We know that,

At point of contact, radius and tangent are perpendicular ∠OTP = 90°

So, ∆OTP is right angled triangle.

Then by Pythagoras theorem, we have

OP2 = OT2 + PT2

262 = OT2 + 102

OT2 = 676 – 100

OT = √576

OT = 24 cm

Thus, OT = length of tangent = 24 cm

**4. If from any point on the common chord of two intersecting circles, tangents be drawn to the circles, prove that they are equal.**

**Solution:**

Let the two circles intersect at points X and Y.

So, XY is the common chord.

Suppose ‘A’ is a point on the common chord and AM and AN be the tangents drawn A to the circle

Then it’s required to prove that AM = AN.

In order to prove the above relation, following property has to be used.

“Let PT be a tangent to the circle from an external point P and a secant to the circle through P intersecting the circle at points A and B, then PT2 = PA × PB”

Now AM is the tangent and AXY is a secant

∴ AM2 = AX × AY … (i)

Similarly, AN is a tangent and AXY is a secant

∴ AN2 = AX × AY …. (ii)

From (i) & (ii), we have AM2 = AN2

∴ AM = AN

Therefore, tangents drawn from any point on the common chord of two intersecting circles are equal.

- Hence Proved

**5. If the quadrilateral sides touch the circle, prove that sum of pair of opposite sides is equal to the sum of other pair.**

**Solution:**

Consider a quadrilateral ABCD touching circle with centre O at points E, F, G and H as shown in figure.

We know that,

The tangents drawn from same external points to the circle are equal in length.

Consider tangents:

1. From point A [AH & AE]

AH = AE … (i)

2. From point B [EB & BF]

BF = EB … (ii)

3. From point C [CF & GC]

FC = CG … (iii)

4. From point D [DG & DH]

DH = DG …. (iv)

Adding (i), (ii), (iii), & (iv)

(AH + BF + FC + DH) = [(AC + CB) + (CG + DG)]

⟹ (AH + DH) + (BF + FC) = (AE + EB) + (CG + DG)

⟹ AD + BC = AB + DC [from fig.]

Therefore, the sum of one pair of opposite sides is equal to other.

- Hence Proved

**6. Out of the two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle.Solution:**

Let C1 and C2 be the two circles having same center O.

And, AC is a chord which touches the C1 at point D

let’s join OD.

So, OD ⊥ AC

AD = DC = 4 cm [perpendicular line OD bisects the chord]

Thus, in right angled ∆AOD,

OA² = AD² + DO² [By Pythagoras theorem]

DO² = 5² – 4² = 25 – 16 = 9

DO = 3 cm

Therefore, the radius of the inner circle OD = 3 cm.

**7. A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ.Solution:**

Given: Chord PQ is parallel tangent at R.

To prove: R bisects the arc PRQ.

Proof:

Since PQ || tangent at R.

∠1 = ∠2 [alternate interior angles]

∠1 = ∠3

[angle between tangent and chord is equal to angle made by chord in alternate segment]

So, ∠2 = ∠3

⇒ PR = QR [sides opposite to equal angles are equal]

Hence, clearly R bisects PQ.

**8. Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A.**

**Solution:**

Given,

AB is a diameter of the circle.

A tangent is drawn from point A.

Construction: Draw a chord CD parallel to the tangent MAN.

So now, CD is a chord of the circle and OA is a radius of the circle.

∠MAO = 90°

[Tangent at any point of a circle is perpendicular to the radius through the point of contact]

∠CEO = ∠MAO [corresponding angles]

∠CEO = 90°

Therefore, OE bisects CD.

[perpendicular from center of circle to chord bisects the chord]

Similarly, the diameter AB bisects all the chords which are parallel to the tangent at the point A.

**9. If AB, AC, PQ are the tangents in the figure, and AB = 5 cm, find the perimeter of ∆APQ.**

**Solution:**

Given,

AB, AC, PQ are tangents

And, AB = 5 cm

Perimeter of **∆**APQ,

Perimeter = AP + AQ + PQ

= AP + AQ + (PX + QX)

We know that,

The two tangents drawn from external point to the circle are equal in length from point A,

So, AB = AC = 5 cm

From point P, PX = PB [Tangents from an external point to the circle are equal.]

From point Q, QX = QC [Tangents from an external point to the circle are equal.]

Thus,

Perimeter (P) = AP + AQ + (PB + QC)

= (AP + PB) + (AQ + QC)

= AB + AC = 5 + 5

= 10 cm.

**10. Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at centre.**

**Solution:**

Consider a circle with centre ‘O’ and has two parallel tangents through A & B at ends of diameter.

Let tangent through M intersect the parallel tangents at P and Q

Then, required to prove: ∠POQ = 90°.

From fig. it is clear that ABQP is a quadrilateral

∠A + ∠B = 90° + 90° = 180° [At point of contact tangent & radius are perpendicular]

∠A + ∠B + ∠P + ∠Q = 360° [Angle sum property of a quadilateral]

So,

∠P + ∠Q = 360° – 180° = 180° … (i)

At P & Q

∠APO = ∠OPQ = 1/2 ∠P ….(ii)

∠BQO = ∠PQO = 1/2 ∠Q ….. (iii)

Using (ii) and (iii) in (i) ⇒

2∠OPQ + 2∠PQO = 180°

∠OPQ + ∠PQO = 90° … (iv)

In ∆OPQ,

∠OPQ + ∠PQO + ∠POQ = 180° [Angle sum property]

90° + ∠POQ = 180° [from (iv)]

∠POQ = 180° – 90° = 90°

Hence, ∠POQ = 90°

**11. In Fig below, PQ is tangent at point R of the circle with center O. If ∠TRQ = 30°, find ∠PRS.**

**Solution:**

Given,

∠TRQ = 30°.

At point R, OR ⊥ RQ.

So, ∠ORQ = 90°

⟹ ∠TRQ + ∠ORT = 90°

⟹ ∠ORT = 90°- 30° = 60°

It’s seen that, ST is diameter,

So, ∠SRT = 90° [ ∵ Angle in semicircle = 90°]

Then,

∠ORT + ∠SRO = 90°

∠SRO + ∠PRS = 90°

∴ ∠PRS = 90°- 30° = 60°

**12. If PA and PB are tangents from an outside point P. such that PA = 10 cm and ∠APB = 60°. Find the length of chord AB.**

**Solution:**

Given,

AP = 10 cm and ∠APB = 60°

Represented in the figure

We know that,

A line drawn from centre to point from where external tangents are drawn divides or bisects the angle made by tangents at that point

So, ∠APO = ∠OPB = 1/2 × 60° = 30°

And, the chord AB will be bisected perpendicularly

∴ AB = 2AM

In ∆AMP,

AM = AP sin 30°

AP/2 = 10/2 = 5cm [As AB = 2AM]

So, AP = 2 AM = 10 cm

And, AB = 2 AM = 10cm

Alternate method:

In ∆AMP, ∠AMP = 90°, ∠APM = 30°

∠AMP + ∠APM + ∠MAP = 180°

90° + 30° + ∠MAP = 180°

∠MAP = 60°

In ∆PAB, ∠MAP = ∠BAP = 60°, ∠APB = 60°

We also get, ∠PBA = 60°

∴ ∆PAB is equilateral triangle

AB = AP = 10 cm

**13. In a right triangle ABC in which ∠B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC at P. Prove that the tangent to the circle at P bisects BC.Solution:**

Let O be the center of the given circle. Suppose, the tangent at P meets BC at Q.

Then join BP.

Required to prove: BQ = QC

Proof :

∠ABC = 90° [tangent at any point of circle is perpendicular to radius through the point of contact]

In ∆ABC, ∠1 + ∠5 = 90° [angle sum property, ∠ABC = 90°]

And, ∠3 = ∠1

[angle between tangent and the chord equals angle made by the chord in alternate segment]

So,

∠3 + ∠5 = 90° ……..(i)

Also, ∠APB = 90° [angle in semi-circle]

∠3 + ∠4 = 90° …….(ii) [∠APB + ∠BPC = 180°, linear pair]

From (i) and (ii), we get

∠3 + ∠5 = ∠3 + ∠4

∠5 = ∠4

⇒ PQ = QC [sides opposite to equal angles are equal]

Also, QP = QB

[tangents drawn from an internal point to a circle are equal]

⇒ QB = QC

– Hence proved.

**14. From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of ∆PCD.Solution:**

Given,

PA and PB are the tangents drawn from a point P outside the circle with centre O.

CD is another tangents to the circle at point E which intersects PA and PB at C and D respectively.

PA = 14 cm

PA and PB are the tangents to the circle from P

So, PA = PB = 14 cm

Now, CA and CE are the tangents from C to the circle.

CA = CE ….(i)

Similarly, DB and DE are the tangents from D to the circle.

DB = DE ….(ii)

Now, perimeter of ∆PCD

= PC + PD + CD

= PC + PD + CE + DE

= PC + CE + PD + DE

= PC + CA + PD = DB {From (i) and (ii)}

= PA + PB

= 14 + 14

= 28 cm

**15. In the figure, ABC is a right triangle right-angled at B such that BC = 6 cm and AB = 8 cm. Find the radius of its incircle.**

**Solution:**

Given,

In right ∆ABC, ∠B = 90°

And, BC = 6 cm, AB = 8 cm

Let r be the radius of incircle whose centre is O and touches the sides AB, BC and CA at P, Q and R respectively.

Since, AP and AR are the tangents to the circle AP = AR

Similarly, CR = CQ and BQ = BP

OP and OQ are radii of the circle

OP ⊥ AB and OQ ⊥ BC and ∠B = 90° (given)

Hence, BPOQ is a square

Thus, BP = BQ = r (sides of a square are equal)

So,

AR = AP = AB – BD = 8 – r

and CR = CQ = BC – BQ = 6 – r

But AC² = AB² + BC² (By Pythagoras Theorem)

= (8)² + (6)² = 64 + 36 = 100 = (10)²

So, AC = 10 cm

⇒AR + CR = 10

⇒ 8 – r + 6 – r = 10

⇒ 14 – 2r = 10

⇒ 2r = 14 – 10 = 4

⇒ r = 2

Therefore, the radius of the incircle = 2 cm

**16. Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.Solution:**

Let mid-point of an arc AMB be M and TMT’ be the tangent to the circle.

Now, join AB, AM and MB.

Since, arc AM = arc MB

⇒ Chord AM = Chord MB

In ∆AMB, AM = MB

⇒ ∠MAB = ∠MBA ……(i)

[equal sides corresponding to the equal angle]

Since, TMT’ is a tangent line.

∠AMT = ∠MBA

[angle in alternate segment are equal]

Thus, ∠AMT = ∠MAB [from Eq. (i)]

But ∠AMT and ∠MAB are alternate angles, which is possible only when AB || TMT’

Hence, the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

– Hence proved

**17. From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP = diameter of the circle, show that ∆APB is equilateral.Solution:**

Given: From a point P outside the circle with centre O, PA and PB are the tangents to the circle such that OP is diameter.

And, AB is joined.

Required to prove: APB is an equilateral triangle

Construction: Join OP, AQ, OA

Proof:

We know that, OP = 2r

⇒ OQ + QP = 2r

⇒ OQ = QP = r

Now in right ∆OAP,

OP is its hypotenuse and Q is its mid-point

Then, OA = AQ = OQ

(mid-point of hypotenuse of a right triangle is equidistances from its vertices)

Thus, ∆OAQ is equilateral triangle. So, ∠AOQ = 60°

Now in right ∆OAP,

∠APO = 90° – 60° = 30°

⇒ ∠APB = 2 ∠APO = 2 x 30° = 60°

But PA = PB (Tangents from P to the circle)

⇒ ∠PAB = ∠PBA = 60°

Hence ∆APB is an equilateral triangle.

**18. Two tangents segments PA and PB are drawn to a circle with centre O such that ∠APB = 120°. Prove that OP = 2 AP.Solution:**

Given: From a point P. Outside the circle with centre O, PA and PB are tangents drawn and ∠APB = 120°

And, OP is joined.

Required to prove: OP = 2 AP

Construction: Take mid-point M of OP and join AM, join also OA and OB.

Proof:

In right ∆OAP,

∠OPA = 1/2∠APB = 1/2 (120°) = 60°

∠AOP = 90° – 60° = 30° [Angle sum property]

M is mid-point of hypotenuse OP of ∆OAP [from construction]

So, MO = MA = MP

∠OAM = ∠AOM = 30° and ∠PAM = 90° – 30° = 60°

Thus, ∆AMP is an equilateral triangle

MA = MP = AP

But, M is mid-point of OP

So,

OP = 2 MP = 2 AP

– Hence proved.

**19. If ∆ABC is isosceles with AB = AC and C (0, r) is the incircle of the ∆ABC touching BC at L. Prove that L bisects BC.Solution:**

Given: In ∆ABC, AB = AC and a circle with centre O and radius r touches the side BC of ∆ABC at L.

Required to prove : L is mid-point of BC.

Proof :

AM and AN are the tangents to the circle from A.

So, AM = AN

But AB = AC (given)

AB – AN = AC – AM

⇒ BN = CM

Now BL and BN are the tangents from B

So, BL = BN

Similarly, CL and CM are tangents

CL = CM

But BN = CM (proved aboved)

So, BL = CL

Therefore, L is mid-point of BC.

**20. AB is a diameter and AC is a chord of a circle with centre O such that ∠BAC = 30°. The tangent at C intersects AB at a point D. Prove that BC = BD. [NCERT Exemplar] Solution:**

Required to prove: BC = BD

Join BC and OC.

Given, ∠BAC = 30°

⇒ ∠BCD = 30°

[angle between tangent and chord is equal to angle made by chord in the alternate segment]

∠ACD = ∠ACO + ∠OCD

∠ACD = 30° + 90° = 120°

[OC ⊥ CD and OA = OC = radius ⇒ ∠OAC = ∠OCA = 30°]

In ∆ACD,

∠CAD + ∠ACD + ∠ADC = 180° [Angle sum property of a triangle]

⇒ 30° + 120° + ∠ADC = 180°

⇒ ∠ADC = 180° – (30° + 120°) = 30°

Now, in ∆BCD,

∠BCD = ∠BDC = 30°

⇒ BC = BD [As sides opposite to equal angles are equal]

- Hence Proved

**21. In the figure, a circle touches all the four sides of a quadrilateral ABCD with AB = 6 cm, BC = 7 cm, and CD = 4 cm. Find AD.Solution:**

Given,

A circle touches the sides AB, BC, CD and DA of a quadrilateral ABCD at P, Q, R and S respectively.

AB = 6 cm, BC = 7 cm, CD = 4cm

Let AD = x

As AP and AS are the tangents to the circle

AP = AS

Similarly,

BP = BQ

CQ = CR

and OR = DS

So, In ABCD

AB + CD = AD + BC (Property of a cyclic quadrilateral)

⇒ 6 + 4 = 7 + x

⇒ 10 = 7 + x

⇒ x = 10 – 7 = 3

Therefore, AD = 3 cm.

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

**Solution:**

Given: TS is a tangent to the circle with centre O at P, and OP is joined.

Required to prove: OP is perpendicular to TS which passes through the centre of the circle

Construction: Draw a line OR which intersect the circle at Q and meets the tangent TS at R

Proof:

OP = OQ (radii of the same circle)

And OQ < OR

⇒ OP < OR

similarly, we can prove that OP is less than all lines which can be drawn from O to TS.

OP is the shortest

OP is perpendicular to TS

Therefore, the perpendicular through P will pass through the centre of the circle

– Hence proved.

**23. Two circles touch externally at a point P. From a point T on the tangent at P, tangents TQ and TR are drawn to the circles with points of contact Q and R respectively. Prove that TQ = TR.**

**Solution:**

Given: Two circles with centres O and C touch each other externally at P. PT is its common tangent

From a point T: PT, TR and TQ are the tangents drawn to the circles.

Required to prove: TQ = TR

Proof:

From T, TR and TP are two tangents to the circle with centre O

So, TR = TP ….(i)

Similarly, from point T

TQ and TP are two tangents to the circle with centre C

TQ = TP ….(ii)

From (i) and (ii) ⇒

TQ = TR

– Hence proved.

**24. A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm. AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the ∆ABC.Solution:**

Given: Two tangents are drawn from an external point A to the circle with centre O. Tangent BC is drawn at a point R and radius of circle = 5 cm.

Required to find : Perimeter of ∆ABC.

Proof:

We know that,

∠OPA = 90°[Tangent at any point of a circle is perpendicular to the radius through the point of contact]

OA² = OP² + PA² [by Pythagoras Theorem]

(13)² = 5² + PA²

⇒ PA² = 144 = 12²

⇒ PA = 12 cm

Now, perimeter of ∆ABC = AB + BC + CA = (AB + BR) + (RC + CA)

= AB + BP + CQ + CA [BR = BP, RC = CQ tangents from internal point to a circle are equal]

= AP + AQ = 2AP = 2 x (12) = 24 cm

[AP = AQ tangent from internal point to a circle are equal]

Therefore, the perimeter of ∆ABC = 24 cm.