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

Question

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?

Answer 1

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 $O$.

Answer 2

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 $(x,y)$