Tangents $O P$ and $O Q$ are drawn from origin $O$ to the circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ with centre $C$ . Prove that the centre of circle circumscribing $△ O P Q$ is $(- g / 2, - f / 2)$ and the area of quadrilateral $O P C Q$ is $\sqrt{c (g^{2} + f^{2} - c)}$ .

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Solution

Circumcircle of $△ O P Q$ is $S + λ P = 0$ where $P$ is chord of contact of $O (0, 0)$ .
$(x^{2} + y^{2} + 2 g x + 2 f y + c) + λ (g x + f y + c) = 0$
It passes through $O (0, 0)$
$∴ c + λ c = 0$ or $λ = - 1$
$∴ x^{2} + y^{2} + g x + f y = 0$ is the circumcircle of $△ O P Q$
whose centre is $(- g / 2, - f / 2)$ .
Area of Quadrilateral $= 2 △ O C P = 2 . \frac{1}{2} . r . t$
$= \sqrt{(g^{2} + f^{2} - c)} \sqrt{c} = \sqrt{c (g^{2} + f^{2} - c)}$
$∵ t = \sqrt{S^{'}} = \sqrt{c}$ where $t$ is length of tangent drawn from $(0, 0$ .

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