If $O A$ and $O B$ are the tangents from the origin to the circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0,$ and $C$ is the centre of the circle, the area of the quadrilateral $O A C B$ is

\frac{1}{2} \sqrt{c (g^{2} + f^{2} - c}

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\sqrt{c (g^{2} + f^{2} - c)}

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c \sqrt{(g^{2} + f^{2} - c)}

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\frac{\sqrt{g^{2} + f^{2} - c}}{c}

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Solution

The correct option is C $\sqrt{c (g^{2} + f^{2} - c)}$
Since $O A = O B$ and $C A = C B$

The diagonal $O C$ divides the quadrilateral $O A C B$ in two equal right-angled triangles, $Δ O A C$ and $Δ O B C$ .

Therefore, the area of the quadrilateral $O A C B$ is given as,

$\Rightarrow 2 (a r e a o f t r i a n g l e O A C) = 2 \times (1 / 2) O A \times A C$

$= \sqrt{0 + 0 + 2 g \times 0 + 2 f \times 0 + c} \sqrt{g^{2} + f^{2} - c}$
$= \sqrt{c (g^{2} + f^{2} - c)}$

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