# RD SHARMA Solutions for Class 10 Maths Chapter 8 - Circles

## Chapter 8 - Circles Exercise Ex. 8.1

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} .

## Chapter 8 - Circles Exercise Ex. 8.2

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

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°

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.

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

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

∠AOQ=58° (given)

In right ∆BAT,

∠ABT + ∠BAT + ∠ATB=180°

29° + 90° + ∠ATB=180°

∠ATB = 61°

that is, ∠ATQ = 61°

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)

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

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^{˚}

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)

## Chapter 8 - Circles Exercise 8.48

radius = 5 cm

So, OP = 5 cm

OQ = 12 cm

So, the correct option is (d).

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).

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).

## Chapter 8 - Circles Exercise 8.49

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).

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).

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).

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).

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).

## Chapter 8 - Circles Exercise 8.50

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).

## Chapter 8 - Circles Exercise 8.51

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).

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).

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).

## Chapter 8 - Circles Exercise 8.52

PA = PD

AQ = QB

and PQ = PA + AQ

PQ = PD + QB

Hence PD + QB = PQ

So, the correct option is (a).

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).

## Chapter 8 - Circles Exercise 8.53

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).

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).

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).

## Chapter 8 - Circles Exercise 8.54

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).

## Chapter 8 - Circles Exercise 8.55

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).

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).

## Chapter 8 - Circles Exercise 8.56

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).

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).

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).

## Chapter 8 - Circles Exercise 8.57

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).

## Chapter 8 - Circles Exercise 8.58

### Other Chapters for CBSE Class 10 Mathematics

Chapter 1- Real Numbers Chapter 2- Polynomials Chapter 3- Pairs of Linear Equations in Two Variables Chapter 4- Quadratic Equations Chapter 5- Arithmetic Progressions Chapter 6- Co-ordinate Geometry Chapter 7- Triangles Chapter 9- Constructions Chapter 10- Trigonometric Ratios Chapter 11- Trigonometric Identities Chapter 12- Heights and Distances Chapter 13- Areas Related to Circles Chapter 14- Surface Areas and Volumes Chapter 15- Statistics Chapter 16- Probability### RD SHARMA Solutions for CBSE Class 10 Subjects

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