Question

Let $y=x+2,4 y=3 x+6$ and $3 y=4 x+1$ be three tangent lines to the circle $(x-h)^2+(y- k )^2= r ^2$. Then $h + k$ is equal to :

• A. 5
• B. $5(1+\sqrt{2})$
• C. 6
• D. $5 \sqrt{2}$

Answer 1

Answer: (A)

$L _1: y = x +2, L _2: 4 y =3 x +6, L _3: 3 y =4 x +1$
Bisector of lines $L _2 \& L _3$
$\frac{4 x-3 y+1}{5}=\pm\left(\frac{3 x-4 y+6}{5}\right)$
$(+) 4 x-3 y+1=3 x-4 y+6$
$x+y=5$
Centre lies on Bisector of $4 x-3 y+1=0 \&$
(0) $3 x-4 y+6=0$
$\Rightarrow h + k =5$