Let a circle $C$ touch the lines $L_1:4x-3y+K_1=0$ and $L_2:4x-3y+K_2=0,{ K}_1,{ K}_2\in \mathit{R}$. If a line passing through the centre of the circle $C$ intersects $L_1$ at $\left(-1,2\right)$ and $L_2$ at $\left(3,-6\right)$, then the equation of the circle $C$ is

Given,

$L_1=4x-3y+K_1=0; L_2=4x-3y+K_2=0$

Here line $L_1$ and $L_2$ are parallel.

So,

Now point $A\left(-1,2\right)$ will satisfy $L_1$ so, $-4-3\times 2+K_2=0$ $\Rightarrow K_2=10$

Also point $B\left(3,-6\right)$ will satisfy $L_2=4x-3y+K_2$

So $4\times 3-3\times \left(-6\right)+K_2=0 \Rightarrow K_2=-30$

Now distance between the parallel line will be the diameter

Diameter $=\left|\frac{K_1-K_2}{\sqrt{4^2+3^2}}\right|$

$=\left|\frac{10+30}{5}\right|=8$

So radius $\frac{8}{2}=4$

Now mid-point of $AB$ will give us centre of circle by symmetry, so by midpoint formula in $AB$ we get centre $\left(1,-2\right)$,

Now equation of circle will be ${\left(x-1\right)}^2+{\left(y+2\right)}^2=4^2$