Question

Let $O$ be the origin, $OP$ and $OQ$ be two perpendicular chords of equal length of the circle $x^{2}+y^{2}-4 x+8 y=0$. Let $m_{1}$ and $m_{2}$ be the slopes of chords $OP$ and $OQ$.
Evaluate $\left|\frac{m_{1}}{m_{2}}\right|$, where $m_{1}>m_{2}$.

Answer 1

$x^{2}+y^{2}-4 x+8 y=0$
$\Leftrightarrow(x-2)^{2}+(y+4)^{2}=20$
Centre $C =(2,-4)$
Slope of $OC =\frac{-4-0}{2-0}=-2$
Let OP have slope $m .$ Then
$\tan 45^{\circ}=\left|\frac{m+2}{1-2 m}\right|$
$\Leftrightarrow m+2=\pm(1-2 m)$
$\Leftrightarrow 3 m=-1$ or $m=3$
$\Rightarrow m_{1}=3, m_{2}=-\frac{1}{3}$
$\Leftrightarrow\left|\frac{m_{1}}{m_{2}}\right|=\left|\frac{3}{\left(-\frac{1}{3}\right)}\right|=9$