Show that pair of straight lines $ax^{2}+2hxy+ay^{2}+2gx+2fy+c=0$ meet coordinate axes in concyclic points. Also find equation of the circle Show that pair of straight lines $ax^{2}+2hxy+ay^{2}+2gx+2fy+c=0$ meet coordinate axes in concyclic points. Also find equation of the circle through those cyclic points
My Attempt:
Given equation to the pair of straight lines is
$ax^2+2hxy+ay^2+2gx+2fy+c=0$
Let the lines be
$l_1x+m_1y+n_1=0$
and 
$l_2x+m_2y+n_2=0$
Now what should I do next?
 A: Put $x=0$, $$ay^2+2fy+c=0$$
$$y_{1}y_{2} = \frac{c}{a}$$
Put $y=0$, $$ax^2+2gx+c=0$$
$$x_{1}x_{2} = \frac{c}{a}$$
Hence, $$x_{1}x_{2} = y_{1}y_{2}$$
By Converse of Intersecting chord theorem, the intercepts are concyclic.

Note briefly:
Note that it's also true for any non-degenerate conics such that $g^2>ac$,  $f^2>ac$ and $c\neq 0$.
For two straight lines,
$$\begin{vmatrix}
    a & h & g \\
    h & a & f \\
    g & f & c
  \end{vmatrix}=0$$

Now the equation of the required circle is
$$\fbox{$a(x^2+y^2)+2(fx+gy)+c=0$}$$
See the link here.
In the diagram below, $A=(x_{1},0), B=(x_{2},0), C=(0,y_{1}), D=(0,y_{2})$, $AB$ and $CD$ are the intersecting chords that meet at the origin $O$.

A: Begin by finding intercepts:
$$ax^2+2hxy+ay^2+2gx+2fy+c=0$$
For x intercepts,
$$ ay^2 + 2fy +c = 0$$
For y intercepts,
$$ ax^2 + 2gx + c = 0$$
Let the points corresponding to root be: $(a,0) , (b,0) , (0,c) , (0,d)$
From this answer here, the equation of circle through these points:
$$
   \det\begin{bmatrix}
     x^2 +y^2 & x & y & 1 \\
     b^2 & b & 0 & 1 \\
     c^2 & 0 & c& 1 \\
    d^2 & 0 & d& 1 \\
   \end{bmatrix} = 0
$$
For checking if $(a,0)$ is on this circle , simply plug it in place of $(x,y)$
