$$ \begin{aligned} & \text { If the } \\ & a x^2+2 h x y+b y^2+2 g x+2 f y+c=0 \end{aligned} \text {…
$$
\begin{aligned}
& \text { If the } \\
& a x^2+2 h x y+b y^2+2 g x+2 f y+c=0
\end{aligned} \text { equation }
$$
represents a pair of straight lines, then the square of the distance of their point of intersection from the origin is
$\frac{c(a+b)-a f^2-b g^2}{a b-h^2}$
$\frac{c(a+b)+f^2+g^2}{a b-h^2}$
$\frac{c(a+b)-f^2-g^2}{a b-h^2}$
$\frac{c(a+b)-f^2-g^2}{\left(a b-h^2\right)^2}$
Solution
We know that the point of intersection of the pair of straight line is
$
\left(\sqrt{\frac{f^2-b c}{h^2-a b}}, \sqrt{\frac{g^2-a c}{h^2-a b}}\right)
$
Required distance
$
=\left[\sqrt{\left(\sqrt{\frac{f^2-b c}{h^2-a b}}-0\right)^2+\left(\sqrt{\frac{g^2-a c}{h^2-a b}}-0\right)^2}\right]^2
$
$\begin{aligned} & =\frac{f^2-b c}{h^2-a b}+\frac{g^2-a c}{h^2-a b} \\ & =\frac{f^2+g^2-c(a+b)}{h^2-a b} \\ & =\frac{c(a+b)-f^2-g^2}{a b-h^2}\end{aligned}$