## Circles Exercise Ex. 8A (By:- Ajay Raj)

### 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

### 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.

And OBP = 90

So, OAP = OBP = 90

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

### 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

### 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

AB = 10 cm

Also, BP and BQ are tangents to the circle

Further, CQ and CR are tangents to the circle

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,