The chord of contact of tangents from a point $P$ to a circle passes through $Q$ . If $l_{1}$ and $l_{2}$ are the lengths of the tangents from $P$ and $Q$ to the circle, then $P Q$ is equal to

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$\frac{l_{1} + l_{2}}{2}$

$\frac{l_{1} - l_{2}}{2}$

$\sqrt{{l_{1}}^{2} + l_{2}^{2}}$

$\sqrt{{l_{1}}^{2} - l_{2}^{2}}$

answer is C.

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Detailed Solution

Let $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ be two points and

$x^{2} + y^{2} = a^{2}$ be the given circle. Then, the chord of contact 0f tangents drawn from $P$ to the given circle is

$x x_{1} + y y_{1} = a^{2}$ It will pass through $Q (x_{2}, y_{2})$ if

$x_{1} x_{2} + y_{1} y_{2} = a^{2}$
Now, $l_{1} = \sqrt{x_{1}^{2} + y_{1}^{2} - a^{2}}, l_{2} = \sqrt{x_{2}^{2} + y_{2}^{2} - a^{2}}$

$∴ P Q = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

$\Rightarrow P Q = \sqrt{(x_{2}^{2} + y_{2}^{2}) + (x_{1}^{2} + y_{1}^{2}) - 2 (x_{1} x_{2} + y_{1} y_{2})}$

$\begin{array}{l} \Rightarrow P Q = \sqrt{(x_{2}^{2} + y_{2}^{2}) + (x_{1}^{2} + y_{1}^{2}) - 2 a^{2}} [U sing (i)] \\ \Rightarrow P Q = \sqrt{(x_{1}^{2} + y_{1}^{2} - a^{2}) + (x_{2}^{2} + y_{2}^{2} - a^{2})} \\ \Rightarrow P Q = \sqrt{l_{1}^{2} + l_{2}^{2}} \end{array}$