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Let a circle $C$ touch the lines $L_{1}: 4 x-3 y+K_{1}$ $=0$ and $L _{2}: 4 x -3 y + K _{2}=0, K _{1}, K _{2} \in R$. If a line passing through the centre of the circle $C$ intersects $L _{1}$ at $(-1,2)$ and $L _{2}$ at $(3,-6)$, then the equation of the circle $C$ is
$(x-1)^{2}+(y-2)^{2}=4$
$(x+1)^{2}+(y-2)^{2}=4$
$(x-1)^{2}+(y+2)^{2}=16$
$(x-1)^{2}+(y-2)^{2}=16$
Solution
$L _{1}: 4 x -3 y + K _{1}=0$
$L _{2}: 4 x -3 y + K _{2}=0$
now
$-4-6+ K _{1}=0 \Rightarrow K _{1}=10$
$12+18+ K _{2}=0 \Rightarrow K _{2}=-30$
$\Rightarrow \quad$ Tangent to the circle are
$\quad 4 x -3 y +10=0$
$\quad 4 x -3 y -30=0$
Length of diameter $2 r=\frac{|10+30|}{5}=8$ $\Rightarrow r =4$
Now centre is mid point of $A$ \& $B$ $x =1, y =-2$
Equation of circle
$( x -1)^{2}+( y +2)^{2}=16$
