# In Fig. 10.58, there are two concentric circles with centre O of radii 5 cm and 3 cm. from an external point P, tangents PA and PB are drawn to these circles. If AP = 12 cm, find the length of BP.

Given: OA = 5 cm

OB = 3 cm

AP = 12 cm

To find: The length of BP.

Theorem Used:

1)Tangent to a circle at a point is perpendicular to the radius through the point of contact.

2) Pythagoras theorem:

In a right-angled triangle, the squares of the hypotenuse is equal to the sum of the squares of the other two sides.

Explanation:

PA and PB are the tangent drawn from the external point A and B to the radius OA and OB.

From the theorem (1) stated above,

OAP = 90°

OBP = 90°

OA=5cm, OB=3cmandAP=12cm
In
OAP by the theorem (2),

OP2 = OA2 + AP2

OP2 = 52 + 122

OP2 = 25 + 144

OP2 = 169

OP = √169

OP = 13 cm

In OBP by the theorem (2),

OP2 = PB2 + OB2

132 = PB2 + 32

169 = PB2 + 9

160 = PB2

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