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Tangents are drawn from the point $P (6, 8)$ to the circle $x^{2} + y^{2} = r^{2}$ . If the area enclosed by the tangents
and the chord of contact is maximum, then $r =$

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$\frac{15}{2}$

answer is C.

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Detailed Solution

$O P = \sqrt{6^{2} + 8^{2}} = 10$

$P A = \sqrt{100 - r^{2}}$

Area of $Δ P A B = \frac{1}{2} P A \cdot P B \sin 2 θ$

$\begin{array}{l} = \frac{1}{2} P A^{2} \sin 2 θ \\ = (100 - r^{2}) \sin θ \cos θ \end{array}$

$f (r) = {(100 - r^{2})}^{\frac{3}{2}} \frac{r}{100}$

Now, $f^{'} (r) = 0$

$\begin{array}{l} \Rightarrow {(100 - r^{2})}^{\frac{3}{2}} - 3 r^{2} \sqrt{100 - r^{2}} = 0 \\ \Rightarrow 100 - r^{2} = 3 r^{2} \Rightarrow r = 5 \end{array}$

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