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