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Q.

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

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a

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

b

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

c

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

d

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

answer is C.

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Detailed Solution

$\begin{array}{l} P = (x_{1}, y_{1}) and Q = (x_{2}, y_{2}) \\ Equation of tangent to the circle S = 0 at P = (x_{1}, y_{1}) is {xx}_{1} + {yy}_{1} = a^{2} \\ Since it passes through Q (x_{2}, y_{2}) \\ x_{1} x_{2} + y_{1} y_{2} = a^{2} ⟶ (1) \\ Now, l_{1} = \sqrt{x_{1}^{2} + y_{1}^{2} - a^{2}}, l_{2} = \sqrt{x_{2}^{2} + y_{2}^{2} - a^{2}} \\ and PQ = \sqrt{l_{1}^{2} + l_{2}^{2}} \end{array}$

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