The square of the distance from the origin to the point of intersection of the pair of lines $a x^2+2 h x…
The square of the distance from the origin to the point of intersection of the pair of lines $a x^2+2 h x y-a y^2+2 g x+$ $2 \mathrm{fy}+\mathrm{c}=0$ is
The equation of pair of lines is
$
a x^2+2 h x y-a y^2+2 g x+2 f y+c=0.....(i)
$
Let the lines represented by the given equation be
$
\begin{aligned}
& l x+m y+n=0.....(ii) \\
& l^{\prime} x+m^{\prime} y+n^{\prime}=0.....(iii)
\end{aligned}
$
from eqs. (i), (ii) \& (iii)
$
\begin{aligned}
& a x^2+2 h x y-a y^2+2 g x+2 f y+c=(l x+m y+n) \\
& \left(l^{\prime} x+m^{\prime} y+n^{\prime}\right)
\end{aligned}
$
Equating the coefficients, we get
$
\begin{aligned}
& \mathrm{ll}^{\prime}=\mathrm{a}, \mathrm{mm}^{\prime}=-\mathrm{a}, \mathrm{nn}^{\prime}=\mathrm{c} \\
& \mathrm{mn}^{\prime}+\mathrm{m}^{\prime} \mathrm{n}=2 \mathrm{f}, \mathrm{nl}^{\prime}+\mathrm{n}^{\prime} \mathrm{l}=2 \mathrm{~g}, \mathrm{~lm}^{\prime}+\mathrm{ml}^{\prime}=2 \mathrm{~h}
\end{aligned}
$
Solving eqn. (ii) \& (iii) by cross multiplication:
$
\frac{x}{m^{\prime}-m^{\prime} n}=\frac{y}{n^{\prime}-n^{\prime} 1}=\frac{1}{m^{\prime}-l m^{\prime}}
$
So, the point of intersection $\mathrm{P}$ is
$
\left(\frac{m^{\prime}-m^{\prime} n}{m^{\prime}-l m^{\prime}}, \frac{n^{\prime}-n^{\prime} 1}{m^{\prime}-m^{\prime}}\right)
$
Square of distance from the origin
$
\begin{aligned}
& =\left(\frac{m n^{\prime}-m^{\prime} n}{m l^{\prime}-l m^{\prime}}\right)^2+\left(\frac{n l^{\prime}-n^{\prime} l}{m l^{\prime}-l m^{\prime}}\right)^2 \\
& =\frac{\left(m n^{\prime}-m^{\prime} n\right)^2+\left(n l^{\prime}-n^{\prime} l\right)^2}{\left(m l^{\prime}-l m^{\prime}\right)^2} \\
& =\frac{\left(m n^{\prime}+m^{\prime} n\right)^2-4 m m^{\prime} n n^{\prime}+\left(n l^{\prime}+n^{\prime} l\right)^2-4 n^{\prime} l^{\prime}}{\left(m l^{\prime}+l m^{\prime}\right)^2-4 m m^{\prime} l l^{\prime}} \\
& =\frac{4 f^2-4(-a) c+4 g^2-4 a c}{4 h^2-4 a(-a)} \\
& =\frac{f^2+g^2}{a^2+h^2}
\end{aligned}
$