Question

Let a circle $C$ touch the lines $L_1:4x-3y+K_1$ $=0$ and $L_2:4x-3y+K_2=0,K_1,K_2\in R$. 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)

$L_1:4x-3y+K_1=0$
$L_2:4x-3y+K_2=0$
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
$4x-3y+10=0$
$4x-3y-30=0$
Length of diameter $2r=\frac{|10+30|}{5}=8$
$\Rightarrow r=4$
Now centre is mid point of $A$ & $B$
$x=1,y=-2$
Equation of circle
$(x-1{)}^2+(y+2{)}^2=16$