Question

The line $x-2y-1=0$ intersects the circle $x^2+y^2+4x-2y-5=0$ at the points $P$ and $Q$, then $\sqrt{5}PQ$ is

Answer 1

Given circle $x^2+y^2+4x-2y-5=0$
$\Rightarrow (x+2{)}^2+(y-1{)}^2=10$
Now $r=\sqrt{10}$ and center $C(-2,1)$
length of perpendicular from $C$ on line $x-2y-1=0$ will be
$d=\frac{|-2-2-1|}{\sqrt{5}}=\sqrt{5}$
Hence $PQ=$ length of the chord
$\therefore PQ=2\sqrt{10-5}=2\sqrt{5}$ units