If the equation $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ represents a pair of straight lines, then the square of the distance of their point of intersection from the origin is

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$\frac{c (a + b) - f^{2} - g^{2}}{{(a b - h^{2})}^{2}}$

$\frac{c (a + b) - f^{2} - g^{2}}{a b - h^{2}}$

$\frac{c (a + b) - a f^{2} - b g^{2}}{a b - h^{2}}$

$\frac{c (a + b) + f^{2} + g^{2}}{a b - h^{2}}$

answer is C.

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Detailed Solution

Given Equation $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$

$\begin{array}{l} Δ = 0 \Rightarrow a b c + 2 f g h - a f^{2} - b h^{2} - c h^{2} = 0 \\ \Rightarrow 2 f g h = a f^{2} + b g^{2} + c h^{2} - a b c \\ P . I = (\frac{h f - b g}{a b - h^{2}}, \frac{g h - a f}{a b - h^{2}}) a n d O (0, 0) \\ O P^{2} = {(\frac{h f - b g}{a b - h^{2}})}^{2} + {(\frac{g h - a f}{a b - h^{2}})}^{2} \end{array}$

$\begin{array}{l} = \frac{h^{2} f^{2} + b^{2} g^{2} + g^{2} h^{2} + a^{2} f^{2} - 2 f g h (a + b)}{{(a b - h^{2})}^{2}} \\ = \frac{h^{2} f^{2} + b^{2} g^{2} + g^{2} h^{2} + a^{2} f^{2} - (a + b) (a f^{2} + b g^{2} + c h^{2} - a b c)}{{(a b - h^{2})}^{2}} \\ = \frac{f^{2} (h^{2} + a^{2} - a^{2} - a b) + g^{2} (b^{2} + h^{2} - a b - b^{2}) - c (a + b) (h^{2} - a b)}{{(a b - h^{2})}^{2}} \\ \frac{f^{2} (h^{2} - a b) + g^{2} (h^{2} - a b) - c (a + b) (h^{2} - a b)}{{(a b - h^{2})}^{2}} \\ = \frac{c (a + b) - f^{2} - g^{2}}{a b - h^{2}} \end{array}$