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

$L_1:4x-3y+K_1$= 0  ${\mathrm{L}}_2:4\mathrm{x}-3\mathrm{y}+{\mathrm{K}}_2=0$

now

$$\begin{gathered} -4-6+{\mathrm{K}}_1=0\Rightarrow {\mathrm{K}}_1=10 \\ 12+18+{\mathrm{K}}_2=0\Rightarrow {\mathrm{K}}_2=-30 \\ \Rightarrow \text{ Tangent to the circle are } \\ 4 x-3 y+10=0 \\ 4 x-3 y-30=0 \\ \text{ Length of diameter }2r=\frac{|10+30|}{5}=8 \\ \Rightarrow r=4 \end{gathered}$$

Now centre is mid point of A & B

x=1, y= -2

Equation of circle
$(x-1{)}^2+(y+2{)}^2=16$