R S Aggarwal and V Aggarwal Solutions for Class 10 Maths CBSE Chapter 8 Circles

Class 10^th R. S. Aggarwal Maths Solution

CBSE Class 10 Maths

R. S. Aggarwal Solution

Circles Exercise Ex. 8A

Solution 1

Let O be the centre of the circle and PT be the length of tangent from P to the circle.

Then, OT = 20 cm and OP = 29 cm

Since tangent to a circle is always perpendicular to the radius through the point of contact.

∴ ∠OTP = 90^o

In right ΔOTP,

OP² = OT² + PT²

⇒29² = 20² + PT²

⇒PT² = 29² - 20² = (29 - 20)(29 + 20) = 9 × 49 = 441

⇒PT = 21 cm

Hence, the length of the tangent drawn from P to the circle is 21 cm.

Solution 2

PA is the tangent to the circle with centre O and radius, such that PO = 25 cm, PA = 24 cm

In PAO, A = 90,

By Pythagoras theorem:

Hence, the radius of the circle is 7 cm.

Solution 3

Solution 4

Solution 5

Given AP is a tangent at A and OA is radius through A and PA and PB are the tangent segments to circle with centre O.

Therefore, OA is perpendicular to AP, similarly, OB is perpendicular to BP.

OAP = 90

And OBP = 90

So, OAP = OBP = 90

OBP + OAP = (90 + 90) = 180

Thus, the sum of opposite angles of quad. AOBP is 180

AOBP is a cyclic quadrilateral

Solution 6

Solution 7

Given: From an external point P, tangent PA and PB are drawn to a circle with centre O. CD is the tangent to the circle at a point E and PA = 14cm.

Since the tangents from an external point are equal, we have

PA = PB,

Also, CA = CE and DB = DE

Perimeter of PCD = PC + CD + PD

=(PA - CA) + (CE + DE) +(PB - DB)

= (PA - CE) + (CE + DE) + (PB - DE)

= (PA + PB) = 2PA = 2(14) cm

= 28 cm

Hence, the Perimeter of PCD = 28 cm

Solution 8

A circle is inscribed in a triangle ABC touching AB, BC and CA at P, Q and R respectively.

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

AR, AP are the tangents to the circle

AP = AR = 7cm

AB = 10 cm

BP = AB - AP = (10 - 7)= 3 cm

Also, BP and BQ are tangents to the circle

BP = BQ = 3 cm

Further, CQ and CR are tangents to the circle

CQ = CR = 5cm

BC = BQ + CQ = (3 + 5) cm = 8 cm

Hence, BC = 8 cm

Solution 9

Let the circle touches the sides AB, BC, CD and DA at P, Q, R, S respectively

We know that the length of tangents drawn from an exterior point to a circle are equal

AP = AS ----(1) {tangents from A}

BP = BQ ---(2) {tangents from B}

CR = CQ ---(3) {tangents from C}

DR = DS----(4) {tangents from D}

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

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

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

AB + CD = AD + BC

AD = (AB + CD) - BC = {(6 + 4) - 7} cm = 3 cm

Hence, AD = 3 cm

Solution 10

Solution 11

Given 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. PA = 10cm. Join OA, OB and OP.

Then, OB = 4 cm, OA= 6 cm and PA = 10 cm

In triangle OAP,

Hence, BP = 10.9 cm

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Circles Exercise Ex. 8B

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

PM and PN are tangents to a circle.

∴ OM ⏊ PM and ON ⏊ PN

⇒ ∠OMP = 90^o and ∠ONP = 90^o

In ΔOMP and ΔONP,

OM = ON (radii of same circle)

∠OMP = ∠ONP = 90^o

OP = OP (common)

∴ ΔOMP ≌ ΔONP (RHS congruency)

∴ ∠OPM = ∠OPN (CACT)

⇒ ∠OPM

Hence, the length of OP is .

Circles Exercise MCQ

Solution 1

Correct option : (b)

We can draw only 2 tangents from an external point to a circle.

Solution 2

Solution 3

Solution 4

Correct option: (d)

The diameter of the circle always passes through the centre. This means all the diameters of a given circle will intersect at the centre, and hence they cannot be parallel.

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

Solution 40

Solution 41

Solution 42

Solution 43

Solution 44

Solution 45

Solution 46

Solution 47

Solution 48

Solution 49

Solution 50

Correct option: (d)

Options (a), (b) and (c) are all true.

However, option (d) is false since we can draw only parallel tangents on either side of the diameter, which would be parallel to a given line.

Solution 51

Correct option: (d)

Options (a), (b) and (c) are all true.

However, option (d) is false since a straight line can meet a circle at two points even as shown below.

Solution 52

Correct option: (d)

Options (a), (b) and (c) are true.

However, option (d) is false since it is not possible to draw a tangent from a point inside a circle.

Solution 53

Solution 54

Solution 55

Circles Exercise Test Yourself

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

A line intersecting a circle in two distinct points is called a secant.
A circle can have two parallel tangents at the most.
The common point of a tangent to a circle and the circle is called the point of contact.
A circle can have infinitely many tangents.