In the figure, there are two concentric circles with centre $ O $ of radii $ 5 \mathrm{~cm} $ and $ 3 \mathrm{~cm} $. From an external point $ P $, tangents $ P A $ and $ P B $ are drawn to these circles. If $ A P=12 \mathrm{~cm} $, find the length of $ B P $."

Given:

In the figure, there are two concentric circles with centre \( O \) of radii \( 5 \mathrm{~cm} \) and \( 3 \mathrm{~cm} \). From an external point \( P \), tangents \( P A \) and \( P B \) are drawn to these circles.

\( A P=12 \mathrm{~cm} \).

To do:

We have to find the length of \( B P \).

Solution:

From the figure,

$AP = 12\ cm$

In right angled triangle $OAP$,

By Pythagoras theorem,

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

$= 5^2 + (12)^2$

$= 25 + 144$

$= 169$

$= (13)^2$

$\Rightarrow OP = 13\ cm$

In right angled triangle $OBP$,

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

$(13)² = 3^2 + BP^2$

$169 = 9 + BP^2$

$BP^2 = 169 - 9$

$= 160$

$= 16 \times 10$

$\Rightarrow BP = \sqrt{16 \times 10}$

$= 4\sqrt{10}\ cm$

The length of \( B P \) is $ 4\sqrt{10}\ cm$.

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Updated on: 10-Oct-2022

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