Tangents PA and PB are drawn from an external point P to two concentric circles with center O and radii 8 cm and 5 cm respectively, as shown in the figure. If AP$=15$ cm, then find the length of BP.
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Given: Two concentric circles with center O and its tangents AP and BP from external point P. AP$=15$ cm.

To do: To find the value of BP.

Solution: Let us join OP.

$\because$ tangent to a circle is always  $\perp $ to the radius.

$\therefore \angle A=\angle B=90^{o}$

In right triangle $\vartriangle OAP$,

Using pythagoras theorem,

$OP^{2} =OA^{2} +AP^{2}$

$=8^{2} +15^{2}$

$=64+225$

$=289$

$\Rightarrow OP=\sqrt{289} =17\ cm$

In right triangle, $\vartriangle OBP$,

Using pythagoras theorrem,

$OB^{2} +BP^{2} =OP^{2}$

$\Rightarrow 5^{2} +BP^{2} =17^{2}$

$\Rightarrow BP^{2} +25=289$

$\Rightarrow BP^{2} =289-25$

$\Rightarrow BP^{2} =264$

$\Rightarrow BP=\sqrt{2\times 2\times 66}$

$\Rightarrow BP=2\sqrt{66} \ cm$

Therefore the length of $BP=2\sqrt{66} \ cm$

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Updated on: 10-Oct-2022

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