Question

If the lines
$a_{1}x+b_{1}y+c_1=0and{a}_{2}x+b_{2}y+c_2=0$
cut the co-ordinate axes in concyclic points, prove $a_1{a}_2=b_1{b}_2$ and also find the equation of that circle

Answer 1

The curve through the four points of intersection of the given lines and axes whose equation is xy = 0 is
($a_1$ x + $b_1$ y + $c_1$)($a_2$ x + $b_2$ y + $c_2$) + $\lambda$xy = 0 (1)
It represents a circle as the four points are concyclic.
Coeff. of x${}^2$ = Coeff. of y${}^2$
and Coeff. of xy = 0
$\therefore a_1{a}_2=b_1{b}_2$ and also $(a_1{b}_2+a_2{b}_1)$ + $\lambda$ = 0
$\therefore \, \, \, \lambda \, = \, -(a_1b_2 \, + \, a_2b_1)$.
Putting in (1), we get the required circle.
Another form :- If the two lines cut the co-ordinate axes in four concylic points then product of their intercepts on x-axis is equal to product of their intercepts on y-axis. (Rule)
$\therefore -\frac{c_1}{a_1}\cdot -\frac{c_2}{a_2}=-\frac{c_1}{b_1}\cdot -\frac{c_2}{b_2}\therefore a_1{a}_2=b_1{b}_2$