If $a_1{a}_2=b_1{b}_2$ then the lines $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$ cut the coordinate axes in concyclic points. Then equation of the circle, is

The equation of the second degree curve passing through the points of intersection of the given lines with the coordinate axes is

$\left(a_{1}x+b_{1}y+c_1\right)\left(a_{2}x+b_{2}y+c_2\right)+\lambda xy=0$

This will represent a circle, if

Coefficient of $x^2=$ Coefficient of $y^2$ and, Coeff. of$xy = 0$

$\Rightarrow a_1{a}_2=b_1{b}_2\text{ and }{a}_1{b}_2+a_2{b}_1+\lambda =0$

$\Rightarrow a_1{a}_2=b_1{b}_2\text{ and }\lambda =-\left(a_1{b}_2+a_2{b}_1\right)$

Putting the value of').$\lambda$ in (i), we obtain

$\left(a_{1}x+b_{1}y+c_1\right)\left(a_{2}x+b_{2}y+c_2\right)-\left(a_1{b}_2+a_2{b}_1\right)xy=0$

as the equation of the required circle