Mathematics

In the figure (ii) given below, two concentric circles with centre O are 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 BP.

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Join OA and OB.

OA ⊥ AP. (∵ tangent at a point and radius through the point are perpendicular to each other.)

So, in right angled triangle OAP,

$OP^2 = OA^2 + AP^2 \\[1em] OP^2 = 5^2 + 12^2 \\[1em] OP^2 = 25 + 144 \\[1em] OP^2 = 169 \\[1em] OP = \sqrt{169} \\[1em] OP = 13 \text{ cm}.$

OB ⊥ BP. (∵ tangent at a point and radius through the point are perpendicular to each other.)

So, in right angled triangle OBP,

$OP^2 = BP^2 + OB^2 \\[1em] 13^2 = BP^2 + 3^2 \\[1em] 169 = 9 + BP^2 \\[1em] BP^2 = 160 \\[1em] BP = \sqrt{160} \\[1em] BP = 4\sqrt{10} \text{ cm}.$

Hence, the length of BP = $4\sqrt{10}$ cm.

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