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# 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 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 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.

## 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 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

### 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

### 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

## Circles Exercise FA

### Solution 9

1. A line intersecting a circle in two distinct points is called a secant.
2. A circle can have two parallel tangents at the most.
3. This is since we can draw only parallel tangents on either side of a diameter.
4. The common point of a tangent to a circle and the circle is called the point of contact.
5. A circle can have infinitely many tangents.