Question

# A tangent $PQ$ at a point $P$ of a circle of radius $5cm$ meets a line through the centre $O$ at a point $Q$ so that $OQ=12cm$. Length $PQ$ is :

A

$12cm$

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B

$13cm$

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C

$8.5cm$

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D

$\sqrt{119}cm$

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Solution

## The correct option is D $\sqrt{119}cm$Step 1. From the figure, the line that is drawn from the centre of the given circle to the tangent$PQ$ is perpendicular to $PQ$ So, $OP\perp PQ$Step 2. Using Pythagoras theorem in right angled triangle $∆OPQ$ we have, $O{Q}^{2}=O{P}^{2}+P{Q}^{2}\phantom{\rule{0ex}{0ex}}⇒{12}^{2}={5}^{2}+P{Q}^{2}\phantom{\rule{0ex}{0ex}}⇒144=25+P{Q}^{2}\phantom{\rule{0ex}{0ex}}⇒144-25=P{Q}^{2}\phantom{\rule{0ex}{0ex}}⇒119=P{Q}^{2}\phantom{\rule{0ex}{0ex}}⇒P{Q}^{2}=119\phantom{\rule{0ex}{0ex}}⇒PQ=\sqrt{119}\phantom{\rule{0ex}{0ex}}$Hence option (D) is the correct option.

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