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If the pair of lines $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ are equidistant from the origin, then the condition is

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$f^{2} - g^{2} = c (b f - a g)$

$f^{4} - g^{4} = c (b f^{2} - a g^{2})$

$g^{2} (g^{2} - a c) = f^{2} (f^{2} - b c)$

$g (g - a c) = f (f - b c)$

answer is A, C.

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Detailed Solution

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

represents two lines $l_{1} x + m_{1} y + n_{1} = 0 and l_{2} x + m_{2} y + n_{2} = 0$

$l_{1} l_{2} = a, m_{1} m_{2} = b,, l_{1} m_{2} + l_{2} m_{1} = 2 h$

$l_{1} n_{2} + l_{2} n_{1} = 2 g, m_{1} n_{2} + m_{2} n_{1} = 2 f,, n_{1} n_{2} = c$

$\begin{aligned} \frac{|n_{1}|}{\sqrt{l_{1}^{2} + m_{1}^{2}}} = \frac{|n_{2}|}{\sqrt{l_{2}^{2} + m_{2}^{2}}} \\ \Rightarrow n_{1}^{2} (l_{2}^{2} + m_{2}^{2}) = n_{2}^{2} (l_{1}^{2} + m_{1}^{2}) \end{aligned}$

$\Rightarrow l_{1}^{2} n_{2}^{2} - l_{2}^{2} m_{1}^{2} = m_{2}^{2} n_{1}^{2} - m_{1}^{2} n_{2}^{2}$

$\Rightarrow (l_{1} n_{2} + l_{2} n_{1}) (l_{1} n_{2} - l_{2} n_{1}) = (m_{2} n_{1} + m_{1} n_{2}) (m_{2} n_{1} - m_{1} n_{2})$

$\Rightarrow 2 g (l_{1} n_{2} - l_{2} n_{1}) = 2 f (m_{2} n_{1} - m_{1} n_{2})$

$\Rightarrow g^{2} (4 g^{2} - 4 a c) = f^{2} (4 f^{2} - 4 b c)$

$\Rightarrow f^{4} - g^{4} = b c f^{2} - c a g^{2} = c (b f^{2} - a g^{2})$

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