The area of triangle formed by the pair of tangents drawn from $p (6, 8)$ to the circle $x^{2} + y^{2} = r^{2}$ and chord of contact of p w.r.t $x^{2} + y^{2} = r^{2}$ is maximum then $r = . . . .$

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Solution

$T = \sqrt{x^{2} + y^{2} - r^{2}} = \sqrt{6^{2} + 8^{2} - r^{2}}$
$= \sqrt{100 - r^{2}}$
$cos θ = \frac{T}{P M} = \frac{\sqrt{100 - r^{2}}}{(100)}$
$P M = P T cos θ$
$= \sqrt{100 - r^{2}} . \frac{\sqrt{100 - r^{2}}}{(100)} = \frac{100 - r^{2}}{100}$
$A r e a = P T . M L$
$M L = P T sin θ = \frac{r}{100}$
$A r e a = \frac{(100 - r^{2}) \times r}{100}$
$\frac{d A}{d r} = \frac{100 - 3 r^{2}}{100} = 0$
$\Rightarrow 100 = 3 r^{2}$
$r = \sqrt{\frac{100}{3}}$

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