R S Aggarwal Solutions for Class 10 is a beneficial solution guide prepared by subject experts at BYJU’S to help students clear their doubts. Chapter 8 of class 10 is loaded with various exercises that are board and competitive exam-oriented. The set of questions and answers will guide students in getting acquainted with the concepts. Find more concepts on R S Aggarwal Solutions for Maths designed by experts to help students learn the most efficient way possible. Students can avail the R S Aggarwal Solutions and download the pfd for free.

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### Access Solutions to Maths R S Aggarwal Chapter 8 – Circles

Get detailed solutions for all the questions listed under the below exercises:

## Exercise 8A

**Question 1: A point P is at a distance of 29 cm from the centre of a circle of radius 20 cm. Find the length of the tangent drawn from P to the circle. **

**Solution:**

Let O be the center of circle, OP = 29cm (given)

Radius of a circle = 20 cm

Let T be any point on the circumference of the circle, then PT is the tangent to the circle.

From Figure:

OT is radius and PT is the tangent

This implies, OT ⊥ PT

In right ∆OPT,

By Pythagoras Theorem:

OP² = OT² + PT²

(29)² = (20)² + PT²

841 = 400 + PT²

PT² = 441

or PT = 21

Length of tangent PT is 21 cm

**Question 2: A point P is 25 cm away from the centre of a circle and the length of tangent drawn from P to the circle is 24 cm. Find the radius of the circle.**

**Solution:**

Let O be the center of circle, OP = 25 cm (given)

Let T be the any point on the circle, then PT = 24 cm

To Find: Radius of the circle

From Figure:

OT ⊥ PT

In right ∆OPT,

By Pythagoras Theorem:

OP² = OT² + PT²

25² = OT² + (24)²

625 = OT² + 576

OT = 7

Radius of the circle is 7 cm.

**Question 3: Two concentric circles are of radii 6.5 cm and 2.5 cm. Find the length of the chord of the larger circle which touches the smaller circle.**

**Solution:**

Let O be the center of circle. Draw two concentric circles are of radii 6.5 cm and 2.5 cm, and

AB = Chord of the larger circle which touches the smaller circle at C.

From Figure:

OC = radius = 2.5 cm

OA = 6.5 cm

AC = CB

OC ⊥ AB and OC bisects AB at C.

In right ∆OPT,

By Pythagoras Theorem:

OA² = OC² + AC²

6.5² = 2.5² + AC²

42.25 = 6.25 + AC²

AC = 6

Length of chord of a circle = AB = 2xAC = 2 x 6 = 12 cm.

**Question 4: In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at points D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, find the lengths of AD, BE and CF.**

**Solution:**

AB = 12 cm, BC = 8 cm and AC = 10 cm

To Find: Lengths of AD, BE and CF

AD and AF are tangents to the circle from A.

AD = AF = x

Similarly, BD and BE are tangents to the circle.

BD = BE = y

and CE and CF are tangents to the circle

CE = CF = z

x + y + 12 ……(1)

y + z = 8 …..(2)

z + x = 10 …..(3)

Adding (1), (2) and (3), we get

2(x + y + z) = 12 + 8 + 10 = 30

x + y + z = 15 …..(4)

Subtract (1) from (4): z = 3

Subtract (2) from (4): x = 7

Subtract (3) from (4): y = 5

Therefore,

AD = 7 cm, BE = 5 cm and CF = 3 cm

**Question 5: In the given figure, PA and PB are the tangent segments to a circle with centre O. Show that the points A, O, B and P are concyclic.**

**Solution:**

PA and PB are tangents to a circle with center O (given)

To show: Points A, O, B and P are concyclic.

Since OB ⏊ PB and OA ⏊ AP

∠OBP = ∠OAP = 90°

∠OBP + ∠OAP = 90 + 90 = 180°

[Sum of opposite angles in a quadrilateral is 180°]AOBP is a cyclic quadrilateral, thus A, O, B and P are concyclic.

Hence proved.

**Question 6: In the given figure, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB.**

**Solution:**

In the given figure, chord AB of larger circle of the two concentric circles with centre O, touches the smaller circle at C.

To Prove: AC = CB

Join OC, OA and OB.

Now,

AB is tangent to the smaller circles and OC is the radius.

this implies, OC ⊥ AB.

We have two right angled triangles: ∆OAC and ∆OBC

Here OA = OB = radius of same circle

Side OC = OC = common among both the triangles

∆OAC = ∆OBC (RHS axiom)

By c.p.c.t.

AC = CB

Hence proved.

**Question 7: 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:**

In the given figure,

CD is a tangent at E and

PA = 14 cm.

Now,

PA and PB are tangents from P to the circle.

PA = PB ….(1)

CA and CE are tangents to the circle.

CA = CE ….(2)

Similarly, BD and DE are tangents to the circle.

DB = DE

Now, perimeter of ∆PCD = Sum of all the sides

= PC + CD + PD

= PC + (CE + ED) + PD

[Using (1) and (2)]= (PC + CA) + (BD + PD)

= PA + PB

= 14 + 14

= 28

Therefore, perimeter of ∆PCD is 28 cm

**Question 8: A circle is inscribed in a ∆ABC touching AB, BC and AC are P, Q and R respectively. If AB = 10 cm, AR = 7 cm and CR = 5 cm, find the length of BC. **

**Solution:**

AB = 10 cm, AR = 7 cm and CR = 5 cm

AP and AR are the tangents to the circle

AP = AR = 7 cm

From figure: BP = AB – AP = 10 – 7 = 3 cm

Again, BP and BQ are the tangents to the circle, we have

BQ = BP = 3 cm

Similarly, CQ = CR = 5 cm

Now,

BC = CQ + BQ = 5 + 3 = 8

The length of BC is 8 cm.

**Question 9: In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose three sides are AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.**

**Solution:**

Here: AB = 6 cm, BC = 7 cm and CD = 4 cm

Let P, Q, R and S are 4 points touches the sides of a quadrilateral.

We know, tangents from an external point to a circle are always equal.

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

Now from figure, AP + BP + CR + DR = AS + BQ + CQ + DC

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

AB + CD = AD + BC

6 + 4 = AD + 7

AD = 3

AD = 3 cm. Answer!!

**Question 10: In the given figure, an isosceles triangle ABC, with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC. **

**Solution:**

∆ABC is an isosceles triangle where AB = AC and circumscribed a circle.

The circle touches its sides BC, CA and AB at P, Q and R respectively.

To Prove : P bisects the base BC, i.e. BP = PC

Now form figure,

BR and BP are tangents to the circle.

So, BR = BP …..(1)

AR and AQ are tangents to the circle.

AR = AQ But AB = AC

AB – AR = AC – AQ

BR = CQ …..(2)

Similarly, CP and CQ are tangents to the circle.

CP = CQ ……(3)

From (1), (2) and (3)

BP = PC

Hence Proved.

**Question 11: In the given figure, O is the centre of two concentric circles of radii 4 cm and 6 cm respectively. PA and PB are tangents to the outer and inner circle respectively. If PA = 10 cm, find the length of PB up to one place of decimal.**

**Solution:**

In the given figure,

PA and PB are the tangents drawn from P, to the outer circle and inner circle respectively.

PA = 10 cm

OA and OB are the radii and

PA and PB are two tangents to the circles respectively

So,

OA ⊥ PA and OB ⊥ PB

In right ∆OAP,

By Pythagoras Theorem:

OP² = OA² + PA²

= (6)² + (10)²

OP²= 136 …(1)

From right ∆OBP,

OP² = OB² + PB²

136 = (4)² + PB²

136 = 16 + PB²

[Using equation (1)]We have PB² = 136 – 16 = 120

Or PB = √120 cm = 2√30 cm = 2 x 5.47 = 10.9

Answer: Length of PB is 10.9 cm

**Question 12: In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point of contact D, are of lengths 6 cm and 9 cm respectively. If the area of ∆ABC = 54 cm² then find the lengths of sides AB and AC.**

**Solution:**

In the given figure, ∆ABC circumscribed the circle with centre O.

Radius OD = 3 cm

BD = 6 cm, DC = 9 cm

Area of ∆ABC = 54 cm²

To find : Lengths of AB and AC.

AF and EA are tangents to the circle at point A.

Let AF = EA = x

BD and BF are tangents to the circle at point B.

BD = BF = 6 cm

CD and CE are tangents to the circle at point C.

CD = CE = 9 cm

Now, new sides of the triangle are:

AB = AF + FB = x + 6 cm

AC = AE + EC = x + 9 cm

BC = BD + DC = 6 + 9 = 15 cm

Squaring both sides, we have

54^2 = 54x(x + 15)

x^2 + 15x – 54 = 0

Solve this quadratic equation and find the value of x.

x^2 + 18x – 3x – 54 = 0

x(x + 18) – 3(x + 18) = 0

(x – 3)(x + 18) = 0

Either x = 3 or x = – 18

But x cannot be negative.

So, x = 3

Answer:

AB = x + 6 = 3 + 6 = 9 cm

AC = x + 9 = 3 + 9 = 12 cm

**Question 13: PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangents at P and Q intersect at a point T as shown in the figure. Find the length of TP. **

**Solution**:

Radius of the circle is 3 cm and PQ = 4.8 cm.

PQ is a chord of the circle with centre O.

The tangents at P and Q intersect at point T

So, TP and TQ are tangents

OP and OT are joined.

Join OQ

We have two right triangles: △OPT and △OQT

Here,

OT = OT (Common)

PT = QT (tangents of the circle)

OP = OQ (radius of the same circle)

By Side – Side – Side Criterion

△OPT ≅△OQT

Therefore, ∠POT = ∠OQT

Again, from triangles △OPR and △OQR

OR = OR (Common)

OP = OQ (radius of the same circle)

∠POR = ∠OQR (from above result)

By Side – Angle – Side Criterion

△OPR ≅△OQT

Therefore, ∠ORP = ∠ORQ

Now,

∠ORP + ∠ORQ = 180°

(Sum of linear angles = 180 degrees)

∠ORP + ∠ORP = 180°

∠ORP = 90°

This implies, OR ⏊ PQ and RT ⏊ PQ

Also OR perpendicular from center to a chord bisects the chord,

PR = QR = PQ/2 = 4.8/2 = 2.4 cm

Applying Pythagoras Theorem on right triangle △OPR,

(OP)^2 = (OR)^2 + (PR)^2

(3)^2 = (OR)^2 + (2.4)^2

OR = 1.8 cm

Applying Pythagoras Theorem on right angled △TPR,

(PT)^2 = (PR)^2 + (TR)^2 …(1)

Also, OP ⏊ OT

Applying Pythagoras Theorem on right △OPT,

(PT)^2 + (OP)^2 = (OT)^2

[(PR)^2 + (TR)^2 ] + (OP)^2 = (TR + OR)^2(Using equation (1) and from figure)

(2.4)^2 + (TR)^2 + (3)^2 = (TR + 1.8)^2

4.76 + (TR)^2 + 9 = (TR)^2 + 2(1.8)TR + (1.8)^2

13.76 = 3.6 TR + 3.24

TR = 2.9 cm [approx.]

From (1) => PT^2 = (2.4)^2 + (2.9)^2

PT^2 = 4.76 + 8.41

or PT = 3.63 cm [approx.]

Answer: Length of PT is 3.63 cm.

**Question 14: Prove that the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre.**

**Solution:**

Let O is the centre of the circle.

Let PQ and RS are two parallel tangents which touches the circle at points A and B.

OA and OB are joined.

To Prove : AB passes through point O.

Draw OC || RS ||PQ

Now,

OA = OB = Radius and

PQ is tangent of circle passing through A

OA ⊥ PQ

Which implies, ∠OAP = 90°

RS is the tangent of circle passing through B

OB ⊥ RS

Which implies, ∠OBR = 90°

Since PQ || OC

∠AOC + ∠OAP = 180° (Co-interior angles)

∠AOC + 90° = 180°

∠AOC = 180° – 90° = 90°

Similarly, ∠BOC = 90°

∠AOC + ∠BOC = 90° + 90° = 180°

AOB is a straight line. Therefore, AB passes through the centre of the circle.

**Question 15: In the given figure, a circle with centre O, is inscribed in a quadrilateral ABCD such that it touches the side BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, ∠B = 90° and DS = 5 cm then find the radius of the circle.**

**Solution:**

In the given figure,

O be the centre of circle which is inscribed in a quadrilateral ABCD.

And OP = OQ = r = radius of circle

The circle touches the sides of quadrilateral at P, Q, R and S respectively.

AB = 29 cm, AD = 23 cm, ∠B = 90°

DS = 5 cm

Join OP and OQ.

Now,

OP = OQ = r and ∠B = 90°

So, PBQO is a square.

DR and DS are the tangents to the circle.

DR = DS = 5 cm

AQ and AR are tangents to the circle.

AR = AD – DR = 23 – 5 = 18 cm

AQ = AR = 18 cm

And BQ = AB – AQ = 29 – 18 = 11 cm

Since PBQO is a square.

OP = OQ = BQ = 11 cm

Hence, radius of the circle is 11 cm

**Question 16: In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. If ∠PBT = 30°, prove that BA : AT = 2 : 1.**

**Solution:**

In the given figure,

TP is the tangent from an external point T and

∠PBT = 30°

Now,

∠APB = 90° (Angle in a semicircle)

∠PBT = 30° (given)

So, ∠PAB = 90° – 30° = 60°

But, ∠PAT + ∠PAB = 180° (Linear pair)

∠PAT + 60° = 180°

∠PAT = 180° – 60° = 120°

Also, ∠APT = ∠PBA = 30° (Angles in the alternate segment)

In ∆PAT,

∠PTA = 180° – (120° + 30°) = 180° – 150° = 30°

PA = AT

In right ∆APB,

sin 30^0 = AP/AB

1/2 = AP/AB

AB = 2 AP

Since AP = AT

AB = 2 AT

or AB/AT = 2/1

AB:AT = 2:1

or BA:AT = 2:1. Hence Proved.

## Exercise 8B

**Question 1: In the adjoining figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm, BC = 9 cm and CD = 8 cm. Find the length of side AD.**

**Solution:**

In the given figure,

A circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm, BC = 9 cm and CD = 8 cm

Let P, Q, R and S are points where circle touches the sides AB, BC, CD and DA respectively.

Therefore,

AB + CD = BC + AD

6 + 8 = 9 + AD

14 = 9 + AD

AD = 14 – 9 = 5

Length of AD is 5 cm.

**Question 2: In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 50° then what is the measure of ∠OAB.**

**Solution**:

In the given figure,

∠APB = 50°

PA and PB are two tangents to the circle with centre O.

Join OB.

**Question 3: In the given figure, O is the centre of a circle. PT and PQ are tangents to the circle from an external point P. If ∠TPQ = 70°, find ∠TRQ.**

**Solution:**

Here, O is the centre of circle.

PQ and PT are tangents to the circle from a point P

R is any point on the circle. RT and RQ are joined.

∠TPQ = 70°

Now,

Join TO and QO.

∠TOQ = 180° – 70° = 110°

Here, OQ and OT are perpendicular on QP and TP.

∠TOQ is on the centre and ∠TRQ is on the rest part.

∠TRQ = 1/2∠TOQ = ½( 110°) = 55°

Therefore, ∠TRQ = 55 degrees.

**Question 4: In the given figure, common tangents AB and CD to the two circles with centres O _{1} and O_{2} intersect at E. Prove that AB = CD.**

**Solution:**

In the given figure, common tangents

AB and CD are common tangents to the two circles with centres O_{1} and O_{2} intersecting at point E.

Circle O_{1:}

EA and EC are tangents

EA = EC …(1)

Circle O2:

EB and ED are tangents

EB = ED …(2)

Adding (1) and (2),

EA + EB = EC + ED

AB = CD

Hence proved.

**Question 5: If PT is a tangent to a circle with centre O and PQ is a chord of the circle such that ∠QPT = 70°, then find the measure of ∠POQ.**

**Solution:**

**Question 6: In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 2 cm such that the segments BD and DC into which BC is divided by the point of contact D, are of lengths 4 cm and 3 cm respectively. If the area of ∆ABC = 21 cm² then find the lengths of sides AB and AC.**

**Solution**:

Given:

∆ABC is circumscribed a circle with centre O and radius 2 cm.

Point D divides BC as

BD = 4 cm, DC = 3 cm, OD = 2 cm

Area of ∆ABC = 21 cm²

Join OA, OB, OC, OE and OF.

x = 3.5

Therefore,

AB = AF + FB = 3.5 + 4 = 7.5 cm

AC = AE + CE = 3.5 + 3 = 6.5 cm

**Question 7: Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle (in cm) which touches the smaller circle.**

**Solution:**

Two concentric circles are of radii 5 cm and 3 cm.

Let O be the centre of circle.

Let AB is chord of larger circle which touches the smaller circle at C.

Join OA and OC.

Below is the figure:

**Question 8: Prove that the perpendicular at the point of contact of the tangent to a circle passes through the centre.**

**Solution**:

AB is the tangent to the circle at point P

Let O be the centre.

PL ⊥ AB

**Question 9: In the given figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If ∠PRQ = 120°, then prove that OR = PR + RQ.**

**Solution:**

∠PRQ = ∠QRO = 120°/2 = 60°

RQ and RP are the tangent to the circle.

OQ and OP are radii

OQ ⊥ QR and OP ⊥ PR

Form right ∆OPR,

∠POR = 180° – (90° + 60°) = 30°

and∠QOR = 30°

cos a = PR/OR (suppose ‘a’ be the angle)

cos 60° =PR/OR

1/2 = PR/OR

OR = 2 PR

Again from right ∆OQR,

OR = 2 QR

From both the results, we have

2 PR + 2 QR = 2OR

or OR = PR + RQ

Hence Proved.

**Question 10: In the given figure, a circle inscribed in a triangle ABC touches the sides AB, BC and CA at points D, E and F respectively. If AB = 14 cm, BC = 8 cm and CA = 12 cm. Find the lengths AD, BE and CF.**

**Solution:**

From given statements, let us reframed the given figure:

AB = 14 cm, BC = 8 cm and CA = 12 cm.

**Question 11: In the given figure, O is the centre of the circle. PA and PB are tangents. Show that AOBP is a cyclic quadrilateral. **

**Solution:**

**Question 12: In two concentric circles, a chord of length 8 cm of the larger circle touches the smaller circle. If the radius of the larger circle is 5 cm then find the radius of the smaller circle. **

**Solution:**

Draw a figure based on given instructions:

AC = CB = 8/2 = 4 cm

OA = 5 cm

AB is the tangent and OC is the radius

OC ⊥ AB

**Question 13: In the given figure, PQ is a chord of a circle with centre O and PT is a tangent. If ∠QPT = 60°, find ∠PRQ.**

**Solution**:

**Question 14: In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 60° then find the measure of ∠OAB.**

**Solution:**

From given figure,

PA and PB are the two tangents to the circle.

O be the centre of circle. OA and AB are joined

∠APB = 60°

To Find :∠OAB

∠OAB + 60° = 90°

∠OAB = 90° – 60° = 30°

Therefore, the measure of ∠OAB is 30°.

**Question 15: If the angle between two tangents drawn from an external point P to a circle of radius and centre O, is 60° then find the length of OP.**

**Solution:**

Since, tangents drawn from an external point are equally inclined to the line joining centre to that point.

Draw an image based on given instructions:

Here, PT is the tangent of the circle

OT ⊥ PT

And, ∠TPS = 60° (given)

∠TPO = 60°/2 = 30°

Now, from right ∆TPO:

sin 30° = OT/OP

1/2 = a/OP

or OP = 2a

## R S Aggarwal Solutions for Chapter 8 Circles

In this chapter students will study important concepts on Circles as listed below:

- Circles introduction
- Secant
- Tangent, length of a tangent and number of tangents to a circle.
- Important theorems on circle tangents.

### Key Features of R S Aggarwal Solutions for Class 10 Maths Chapter 8 Circles

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