The lengths of the tangents from the points A and B to the circle are $l_{1} and l_{2}$ respectively. If A and B are conjugate points then ${AB}^{2} =$

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$l_{1}^{2} . l_{2}^{2}$

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

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

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

answer is B.

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Detailed Solution

Let $A (x_{1}, y_{1}) B (x_{2} y_{2})$ conjugate points with respect to

$x^{2} + y^{2} = r^{2} ∴ x_{1} x_{2} + y_{1} y_{2} = r^{2}$

$l_{1} = \sqrt{S_{11}} = \sqrt{x_{1}^{2} + y_{1}^{2} - r^{2}} l_{2} = S_{22} = \sqrt{x_{2}^{2} + y_{2}^{2} - r^{2}}$

${AB}^{2} = {(x_{1} - x_{2})}^{2} + {(y_{2} - y_{1})}^{2} = x_{1}^{2} + y_{1}^{2} + x_{2}^{2} + y_{2}^{2} - 2 (x_{1} x_{2} + y_{1} y_{2})$

$= x_{1}^{2} + y_{1}^{2} + x_{2}^{2} + y_{2}^{2} - 2 r^{2}$

$= (x_{1}^{2} + y_{1}^{2} - r^{2}) + (x_{2}^{2} + y_{2}^{2} - r^{2}) = l_{1}^{2} + l_{2}^{2}$

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