Question

Let a circle $C$ of radius $5$ lie below the $x$-axis. The line $L_1=4x+3y-2$ passes through the centre $P$ of the circle $C$ and intersects the line $L_2:3x-4y-11=0$ at $Q$. The line $L_2$ touches $C$ at the point $Q$. Then the distance of $P$ from the line $5x-12y+51=0$ is

Answer 1

Answer: 11

$4x+3y+2=0$
$3x-4y-11=0$

$\frac{x}{-25}=\frac{y}{50}=\frac{1}{-25}$
$\frac{x-1}{\cos \theta }=\frac{y+2}{\sin \theta }=\pm 5$
$y=-2+5\left(-\frac{4}{5}\right)=-6$
$x=1+5\left(\frac{3}{5}\right)=4$
Req $\cdot$ distance
$\left|\frac{5(4)-12(-6)+51}{13}\right|$
$=\left|\frac{20+72+51}{13}\right|$
$=\frac{143}{13}=11$