# Class 10 R S AGGARWAL AND V AGGARWAL Solutions Maths Chapter 8 - Circles

## 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 Ex. 8A

### Solution 1

PA is the tangent to the circle with center O and radius AO = 8 cm. The point P is at a distance of 17 cm from O.

In PAO, A = 90

By Pythagoras theorem:

Hence, the length of the tangent = 15 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, 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

## Circles Exercise FA

### 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. - This is since we can draw only parallel tangents on either side of a diameter.
- 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.

### Solution 10

### Solution 11

### Solution 12

### Solution 13

### Solution 14

### Solution 15

### Solution 16

### Solution 17

### Solution 18

### Solution 19

### Solution 20