The two conics $a_1{x}^2+2h_{1}xy+b_1{y}^2=c_1$ and

$a_2{x}^2+2h_{2}xy+b_2{y}^2=c_2$ intersect in 4 concyclic points. Then

The equation of a conic passing through the intersection of the given conics is

$\left(a_1{x}^2+2h_{1}xy+b_1{y}^2-c_1\right)+\lambda \left(a_2{x}^2+2h_{2}xy+b_2{y}^2-c_2\right)=0$

This equation will represent a circle, if

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

$\Rightarrow \left(a_1+\lambda u_2\right)=\left(b_1+\lambda b_2\right)\text{ and }2h_1+2\lambda h_2=0$

Eliminating $\lambda$, from these two equations, we get

$\left(a_1-b_1\right)=\frac{h_1}{h_2}\left(a_2-b_2\right)\Rightarrow \left(a_1-b_1\right)h_2=\left(a_2-b_2\right)l_1$