Question

Let $C_{1}$ and $C_{2}$ be the centres of the circles.
$\omega_{1}: x^{2}-2 x+y^{2}=0 $
$\omega_{2}: x^{2}+y^{2}+4 x+8 y+16=0$
respectively. The line joining $C_{1}$ and $C_{2}$ intersects the circles $\omega_{1}$ and $\omega_{2}$ at $Q$ and $P$ respectively. If $(p, q)$ is the centre of the circle drawn on $PQ$ as a diameter, then find $\frac{q}{p}$.

Answer 1

$\omega_{1}: x^{2}-2 x+y^{2}=0$
$\Leftrightarrow(x-1)^{2}+y^{2}=1$
$\Rightarrow$ Centre $C _{1}=(1,0)$, radius $=1$
$\omega_{2}: x^{2}+y^{2}+4 x+8 y+16=0$
$\Leftrightarrow(x+2)^{2}+(y+4)^{2}=4$
$\Rightarrow$ Centre $C _{2}=(-2,-4)$, radius $=2$
$\left| C _{1} C _{2}\right|=\sqrt{(1+2)^{2}+(4)^{2}}=5$

$\Rightarrow| PQ | =\left| C _{1} C _{2}\right|-\left| C _{2} P \right|-\left| C _{1} Q \right| $
$=5-2-1=2$
Let the required circle be $\omega$ with centre $M$.
Radius of circle $\omega=\frac{1}{2}| PQ |=1$
$\Rightarrow \left|C_{2} M\right|=3,\left|C_{1} M\right|=2 $
$\Rightarrow M$ divides $ C_{1} C_{2} $ in $ 2: 3 $ internally.
$M =\left(\frac{2(-2)+3(1)}{5}, \frac{2(-4)+3(0)}{5}\right) $
$=\left(\frac{-1}{5}, \frac{-8}{5}\right)$
$\Rightarrow p=-\frac{1}{5}, q=-\frac{8}{5}$
$\Rightarrow \frac{q}{p}=8$