# Class 10 RD SHARMA Solutions Maths Chapter 8 - Circles

The Central Board of Secondary Education, or CBSE, is a national educational board in India. It sets the curriculum and syllabus and conducts examinations for all affiliated schools. The Board aims to provide a syllabus that lays out a structured framework for teachers and students and thus offers a systematic and comprehensive approach to learning.

The CBSE Class 10 Maths syllabus outlines the specific topics, subtopics, and learning objectives students must cover. It provides a well-rounded mathematical education through chapters such as Real Numbers, Pair of Linear Equations in Two Variables, Quadratic Equations, Arithmetic Progressions, Triangles, Introduction to Trigonometry, Circles, etc. Thus it helps students develop mathematical understanding, problem-solving skills, and logical reasoning abilities.

Circles is an important chapter in Class 10 as it focuses on understanding the properties and characteristics of a circle. It deals with the various elements of the shape, like its radius, diameter, chord, and arc. Studying this chapter lays the foundation for advanced concepts in geometry, which might come into use in various fields like architecture, engineering, and physics.

**RD Sharma Solutions **covers in detail all the topics and concepts included in the CBSE Class 10 Mathematics syllabus. They provide step-by-step solutions to every question covered in the textbook and ensure that the students grasp the underlying principles and techniques. With its simple, crisp and easy-to-understand language, it focuses on simplifying complex concepts so that students can understand and retain them better. The solutions also consist of detailed construction problems related to circles and guide students through the step-by-step construction using a compass and straightedge.

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## Circles Exercise Ex. 8.1

### Solution 1

Fill in the blanks:

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

(ii) A circle may have two parallel tangents.

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

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

(v) The angle between tangent at a point on a circle and the radius through the point is 90^{o} .

### Solution 2

### Solution 3

### Solution 4

## Circles Exercise Ex. 8.2

### Solution 1

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### Solution 17

### Solution 18

### Solution 19

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### Solution 21

### Solution 22

### Solution 23

### Solution 24

AP = AQ , BP = PR and CR = CQ (tangents from an external point)

Perimeter of ∆ABC = AB + BR + RC + CA

= AB + BP + CQ + CA

= AP + AQ

= 2AP

∆APO is a right-angled triangle. AO^{2} = AP^{2} + PO^{2}

13^{2} = AP^{2} + 5^{2}

AP^{2} = 144

AP = 12

∴ Perimeter of ∆ABC = 24 cm

### Solution 25

### Solution 26

### Solution 27

### Solution 28

### Solution 29

### Solution 30

### Solution 31

Since RS is drawn parallel to the tangent PQ,

∠SRQ = ∠PQR

Also, PQ = PR

⇒ ∠PQR = ∠PRQ

In ∆PQR,

∠PQR + ∠PRQ + ∠QPR = 180°

⇒∠PQR + ∠PQR + 30° = 180°

⇒2∠PQR = 150°

⇒∠PQR = 75°

⇒∠SRQ =∠PQR = 75° (alternate angles)

Also, ∠RSQ =∠RQP = 75° (the angle between a tangent and a chord through the point of contact is equal to an angle in the alternate segment.)

In ∆RSQ,

∠RSQ + ∠SRQ + ∠RQS = 180°

⇒75° + 75° + ∠RQS = 180°

⇒ ∠RQS =30°

### Solution 32

### Solution 33

### Solution 34

### Solution 35

### Solution 36

### Solution 37

Since AC is the tangent to the circle with radius 9 cm, we have OB ⊥ AC.

Hence, by applying the Pythagoras Theorem, we have,

OA^{2} = OB^{2} + AB^{2}

⇒ 15^{2} = 9^{2} + AB^{2}

⇒ AB^{2} = 15^{2} - 9^{2}

⇒ AB^{2} = 225 - 81 = 144

∴ AB = 12 cm

We know that the perpendicular drawn from the centre of a circle to any chord of the circle bisects the chord.

Here, OB is the perpendicular and AC is the length of the chord of the circle with radius 15 cm.

So,

AC = 2 × AB = 2 × 12 = 24 cm

Length of the chord of the larger circle which touches the smaller circle = 24 cm.

### Solution 38

### Solution 39

Let PA = PB = x

Tangents drawn from an external point are equal in length. QB = QT = 14 cm , RA = RT = 16 cm

PR = (x + 16) cm, PQ = (x + 14)cm,

QR = 30 cm

= x + 30

Area of ∆PQR

Area of ∆PQR = 336 cm^{2}

Side PR = (12 + 16) = 28 cm

Side PQ = (12 + 14) = 26 cm

### Solution 40

### Solution 41

### Solution 42

Let M and N be the points where AB and AC touch the circle respectively.

Tangents drawn from an external point to a circle are equal

⇒ AM=AN

BD=BM=8 cm and DC=NC=6 cm

### Solution 43

∠AOQ=58° (given)

In right ∆BAT,

∠ABT + ∠BAT + ∠ATB=180°

29° + 90° + ∠ATB=180°

∠ATB = 61°

that is, ∠ATQ = 61°

### Solution 44

### Solution 45

### Solution 46

In ∆AOP,

OA = OP (radii) ∆AOP is an isosceles triangle. OE is a median.

In an isosceles triangle,the median drawn∴∠OEA = 90^{o}

In ∆AOE and ∆ABC,

∠ABC = ∠OEA = 90^{o}

∠A is common.

∆AEO ~ ∆ABC…(AA test)

### Solution 47

### Solution 48

### Solution 49

### Solution 50

PR = PQ…(tangents fromexternal points)

PQ = 5 cm

Also,

OQ is perpendicular to PS …(tangent is perpendicular to the radius)

Now, in a circle,a perpendicular drawn from the centre of a circle bisects the chord.

So, OQ bisects PS.

PQ = QS

QS = 5 cm

PS = 10 cm

### Solution 51

In DPQR,

∠POR is an external angle.

So,

∠POR = ∠PQO + ∠OPQ

Now, PQ is tangent to the circle with radius OQ.

∠PQO = 90^{o}

130˚ = 90˚ + ∠OPQ

∠OPQ = 40^{o}

∠1 = 40^{o}

Now,

Minor arc RT subtends a 130˚ angle at the centre.

So, it will subtend a 65˚angle at any other point on the circle.∠RST = 65˚

∠2 = 65˚

∠1 + ∠2 =105^{˚}

### Solution 52

AP = PB = 12 cm, AC = CQ = 3 cm and QD = DB = 3 cm …(tangent from external point)

PA = 12 cm, PC + CA = 12

PC + 3 = 12

PC= 9 …(i)

Now,

PB = 12

PD + DB = 12

PD + 3 = 12

PD = 9 …(ii)

PC + PD = 18 cm…from (i) and (ii)

## Circles Exercise 8.48

### Solution 1

radius = 5 cm

So, OP = 5 cm

OQ = 12 cm

So, the correct option is (d).

### Solution 2

PQ is a tangent to the circle

So, OP^{2 }+ PQ^{2} = OQ^{2}

OP^{2 }= OQ^{2 }- PQ^{2}

= (25)^{2 }- (24)^{2}

= 49

OP = 7

So, the correct option is (a).

### Solution 3

Given OP = 3 cm

PA = 4 cm

Hence, OA^{2 }= OP^{2} + PA^{2}

OA^{2} = 3^{2 }+ 4^{2}

= 25

OA = 5 cm

So, the correct option is (c).

### Solution 4

## Circles Exercise 8.49

### Solution 5

### Solution 6

### Solution 7

### Solution 8

Tangents from same point to circle have equal length.

Hence Bb = Ba

bC = Cc

Ac = Aa

Let Ba = x then Bb = x

bc = 6 - x and Aa = 8 - x

and Cc = 6 - x and Ac = 8 - x

So AC = AC + cC

= 6 - x + 8 - x

AC = 14 - 2x ......(1)

Also AC^{2 }= AB^{2 }+ BC^{2}

= 8^{2} + 6^{2 }

= 100

AC = 10 .....(2)

from (1) & (2)

14 - 2x = 10

4 = 2x

x = 2 also aB = Ob = radius = 2 cm

So, the correct option is (b).

### Solution 9

### Solution 10

Tangents from same point are of equal length.

AP = AS, PB = BQ

QC = CR, RD = DS

AB = AP + PB .....(1)

BC = BQ + QC ......(2)

CD = CR + RD .....(3)

AD = AS + DS .....(4)

Adding (1) & (3)

AB + CD = AP + BP + CR + RD

= AS + BQ + CQ + DS

= (AS + DS) + (BQ + CQ)

from (2) & (4)

AB + CD = AD + BC

So, the correct option is (b).

### Solution 11

Given OQ = 8 cm

OP = 6 cm

OP^{2 }+ PQ^{2} = OQ^{2}

6^{2 }+ PQ^{2} = 8^{2}

PQ^{2 }= 64 - 36

= 28

PQ =

So, the correct option is (b).

### Solution 12

DA and DC are tangents to circle from same point

so, DA = DC ......(1)

similarly DB = DC ......(2)

(1) + (2)

2DC = DA + DB

2DC = AB

AB = 2 × 4

= 8 cm

So, the correct option is (c).

### Solution 13

AD = AE .......(1)

CD = CF ......(2)

BF = BE .....(3)

from (1)

2AD = 2AE

= AE + AD

= (AB + BE) + (AC + CD)

= AB + BF + AC + CF

= AB + AC + BC

So, the correct option is (b).

### Solution 14

## Circles Exercise 8.50

### Solution 15

### Solution 16

### Solution 17

AP = PQ ....(1)

and OA^{2 }= OP^{2 } + PA^{2}

(15)^{2 }= (9)^{2 }+ AP^{2}

AP^{2 }= 225 - 81

= 144

AP = 12

AP + AQ = 2AP

= 24 cm

So, the correct option is (c).

### Solution 18

### Solution 19

## Circles Exercise 8.51

### Solution 20

AB = 12 cm

BC = 8 cm

AC = 10 cm

Let AD = x

AF = x

BD = 12 - x

and BE = BD = 12 - x

CE = BC - BE

= 8 - (12 - x)

= x - 4

and CE = CF = x - 4

AC = AF + FC

= x + x - 4

AC = 2x - 4

Given, AC = 10 cm

so 2x - 4 = 10

2x = 14

x = 7 cm

AD = 7 cm

So, the correct option is (d).

### Solution 21

AP = BP given

and AP = AQ

also BP = BR

from this, we conclude that

AQ = BR .....(1)

We know CR = CQ .....(2)

from (1) & (2)

AQ + CR = BR + CR

AQ + CQ = BR + CR

AC = BC

So, the correct option is (b).

### Solution 22

AP = 10 cm

AO = 6 cm

OB = 3 cm

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

OP^{2 }= 10^{2} + 6^{2}

OP^{2 }= 136

Also OB^{2 }+ BP^{2} = OP^{2 }

3^{2} + BP^{2 }= 136

BP^{2} = 136 - 9

So, the correct option is (b).

### Solution 23

## Circles Exercise 8.52

### Solution 24

PA = PD

AQ = QB

and PQ = PA + AQ

PQ = PD + QB

Hence PD + QB = PQ

So, the correct option is (a).

### Solution 25

PQ = PT .....(1)

and PT = PR .....(2)

so from (1) & (2)

PQ = PR

PQ = PR = 4.5 cm

QR = PQ + PR

= 4.5 + 4.5 = 9 cm

So, the correct option is (a).

### Solution 26

### Solution 27

## Circles Exercise 8.53

### Solution 28

radius of circle 1 = 3 cm

radius of circle 2 = 5 cm

OP^{2 }= OQ^{2 }+ QP^{2 } and O'S^{2 }+ SR^{2} = O'R^{2 }

OP^{2 }= 4^{2 }+ 3^{2} O'R^{2 }= 5^{2 }+ 12^{2}

= 16 + 9 O'R^{2} = 169

= 25 O'R' = 13 cm

OP = 5 cm

OO' = OK + KO'

= 3 + 5

= 8 cm

PR = PO + OK + KO' + O'R

= 5 + 3 + 5 + 13

= 26 cm

So, the correct option is (b).

### Solution 29

### Solution 30

OB = OC = OA = 5 cm

OQ = OP = 3 cm

OB^{2 }= OQ^{2 }+ BQ^{2}

BQ^{2} = OB^{2 }- OQ^{2}

= 5^{2 }- 3^{2}

= 16

BQ = 4 cm

also BQ = BP

BP = 4 cm

In ΔOPC,

OP^{2} + PC^{2} = OC^{2}

PC^{2 } = OC^{2} - OP^{2}

= 5^{2 }- 3^{2}

= 16

PC = 4 cm

BC = BP + PC = 4 + 4 = 8 cm

So, the correct option is (c).

### Solution 31

Given PR = 7.5 cm

so PR = PQ

PQ = 7.5 cm

PS is the chord to the larger circle. We know that, perpendicular drawn from centre bisect the chords.

Hence PQ = QS

PS = PQ + QS

= 2PQ

= 2 × 7.5

= 15 cm

So, the correct option is (c).

## Circles Exercise 8.54

### Solution 32

AC = AB .....(1)

BD = DP ......(2)

PE = EC ......(3)

AB = 8 so AC = 8 cm

PE = 3 so EC = 3 cm

AE = AC - EC = 8 - 3 = 5 cm

So, the correct option is (c).

### Solution 33

### Solution 34

## Circles Exercise 8.55

### Solution 35

PA = AR

AR = 4 cm

BP = BQ and QC = RC

BQ = 3 cm

Given AC = 11

AR + RC = 11

4 + RC = 11

RC = 7

so QC = 7 cm

BC = BQ + QC

= 3 + 7

= 10 cm

So, the correct option is (b).

### Solution 36

EK = 9 cm

and EK = EM

Hence EM = 9 cm .....(1)

Also EK = ED + DK

and DK = DH

EK = ED + HD .......(2)

EM = EF + FM

and FM = FH

EM = EF + FH ......(3)

(2) + (3)

EK + EM = ED + EF + DH + HF

18 = ED + DF + EF

perimeter = 18 cm

So, the correct option is (a).

### Solution 37

## Circles Exercise 8.56

### Solution 38

AB = 29 cm

AD = 23

DS = 5 cm

DS = DR

so DR = 5 cm

AR = AD - DR

= 23 - 5

= 18 cm

AR = AQ

AQ = 18 cm

BQ = AB - AQ

= 29 - 18

BQ = 11 cm

As OP || BQ and OQ || PB

Hence, OP = BQ

OP = 11 cm

So, the correct option is (a).

### Solution 39

AB = 5 cm

BC = 12 cm

AB^{2 } + BC^{2} = AC^{2}

AC^{2 }= 5^{2} + 12^{2}

= 169

AC = 13 cm

Let BQ = x

AQ = AR = 5 - x

CR = AC - AR

= 13 - (5 - x)

= x + 8

And CP = CR = x + 8

so BP = BC - PC

= 12 - (x + 8)

= 4 - x

But BP = BQ = x

4 - x = x

x = 2

and BQ || OP and OQ || PB

so BQ = PO

PO = 2 cm

So, the correct option is (c).

### Solution 40

### Solution 41

### Solution 42

PT = 3.8 cm

We know

PQ = PT and PT = PR

Hence PQ = 3.8 cm and PR = 3.8 cm

Now, QR = QP + PR

= 3.8 + 3.8

QR = 7.6 cm

So, the correct option is (b).

## Circles Exercise 8.57

### Solution 43

AB = x cm

BC = 7 cm

CR = 3 cm

AS = 5 cm

CR = CQ

CQ = 3 cm

given BC = 7 cm

BQ = BC - QC

= 7 - 3

= 4 cm

And BQ = BP

so BP = 4 cm

Also AS = AP

Hence AP = 5 cm

AB = AP + BP

= 5 + 4

= 9 cm

x = 9 cm

So, the correct option is (b).

### Solution 44

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### Solution 50

## Circles Exercise 8.58

### Solution 51

### Solution 52