Question

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

• A. $(x-1)^{2}+(y-2)^{2}=4$
• B. $(x+1)^{2}+(y-2)^{2}=4$
• C. $( x -1)^{2}+( y +2)^{2}=16$
• D. $( x -1)^{2}+( y -2)^{2}=16$

Answer 1

Answer: (C)

$L _{1}: 4 x -3 y + K _{1}=0$
$L _{2}: 4 x -3 y + K _{2}=0$
now
$-4-6+ K _{1}=0 \Rightarrow K _{1}=10$
$12+18+ K _{2}=0 \Rightarrow K _{2}=-30$
$\Rightarrow $ Tangent to the circle are
$4 x -3 y +10=0$
$4 x -3 y -30=0$
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 $