The radius of the smallest circle passing through the point $(8, 4)$ and cutting the circle $x^{2} + y^{2} = 40$ orthogonally is

\sqrt{5}

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\sqrt{7}

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2 \sqrt{5}

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4 \sqrt{5}

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Solution

The correct option is A $\sqrt{5}$
Let, $x^{2} + 2 g x + 2 f y + c = 0$ be the equation of the circle orthogonal to $x^{2} + y^{2} = 40$ .
Hence, $2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$
$\Rightarrow 0 + 0 = c - 40$
$c = 40$
The point $(8, 4)$ lies on the circle. Hence,
$80 + 16 g + 8 f + 40 = 0$
$10 + 2 g + f + 5 = 0$
$f = (- 15 - 2 g)$
$R a d i u s = \sqrt{g^{2} + f^{2} - 40}$
${R a d i u s}^{2} = g^{2} + f^{2} = 225 + 5 g^{2} + 60 g = 5 (g + 6)^{2} + 45$
To minimise the radius, $g = - 6$
$\Rightarrow R a d i u s = \sqrt{5}$
Hence, option 'A' is correct.

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