Area enclosed by the common tangents of circle $x^{2} + y^{2} = 4$ and ellipse $3 x^{2} + 5 y^{2} = 15$ is

4

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8

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16

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32

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Solution

The correct option is C $16$
Equation of tangent
$y = m x \pm \sqrt{5 m^{2} + 3}$
$\Rightarrow m x - y \pm \sqrt{5 m^{2} + 3} = 0$

Distance from $(0, 0) = 2$
$\Rightarrow 2 = \frac{\sqrt{5 m^{2} + 3}}{\sqrt{1 + m^{2}}}$
Hence $m = 1, - 1$
So four tangents are :
$y = x \pm 2 \sqrt{2} y = - x \pm 2 \sqrt{2}$
The required Area
$= 4 \times \frac{1}{2} \times 2 \sqrt{2} \times 2 \sqrt{2} = 16$

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