Show that the pair of straight lines $ax^2+2hxy+ay^2+2gx+2fy+c=0$ meets the coordinate axes in concyclic points. Also find the.equation of the circle through those cyclic points.
My Approach:
Let $l_1x+m_1y+n_1=0$ and $l_2x+m_2y+n_2=0$ be the two lines represented by $ax^2+2hxy+ay^2+2gx+2fy+c=0$. Now, $$(l_1x+m_1y+n_1)(l_2x+m_2y+n_2l=0$$. Comparing this with $$ax^2+2hxy+ay^2+2gx+2fy+c=0$$. We get: $$l_1l_2=a$$ $$l_1m_2+l_2m_1=2h$$ $$m_1m_2=a$$.
Now, please help me to continue from here.
The given pair of lines meet the $X$ axis at $(x_1, 0), (x_2,0)$, then $x_1, x_2$ are roots of $ax^2 + 2gx + c = 0$ and hence $x_1 + x_2 = -\frac{2g}{a}$ and $x_1x_2 = \frac{c}{a}$. Similarly, they meet the $Y$ axis at $(0,y_1), (0,y_2)$, then $y_1, y_2$ are roots of $ay^2 + 2fy + c = 0$ and hence $y_1 + y_2 = -\frac{2f}{a}$ and $y_1y_2 = \frac{c}{a}$. Now consider the circle $$ax^2 + ay^2 + 2gx + 2fy + c = 0$$ This meets the $X$ axis at points $(x_3,0), (x_4,0)$, then $x_3, x_4$ are roots of $ax^2 + 2gx + c = 0$ and hence $(x_1,0), (x_2,0)$ must coincide with $(x_3, 0), (x_4,0)$ respectively. Similar argument shows that points at which the circle meets the $Y$ axis are the same as those met by the pair of lines. Thus this circle meets the axes at the same points as the pair of lines given and the conclusion follows.