Let a circle C of radius 5 lie below the x-axis.

Question

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

Answer: 11

Accepted Answer

Solving $L_1,L_2$: $Q=(1,-2)$. Since $L_2$ is tangent at $Q$, $PQ=5$ and $P$ lies on $L_1$:
$$(x-1)^2\left(1+\frac{16}9\right)=25\ \Longrightarrow\ (x-1)^2=9\ \Longrightarrow\ x=4,-2.$$

Since $C$ lies below the $x$-axis, $P=(4,-6)$.

$$\text{Distance}=\frac{|20+72+51|}{13}=\frac{143}{13}=11.$$

Solutions · 1