Question

Let $C_1$ and $C_2$ be the centres of the circles.
${\omega }_1:x^2-2x+y^2=0$
${\omega }_2:x^2+y^2+4x+8y+16=0$
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

Answer: 8

${\omega }_1:x^2-2x+y^2=0$
$\iff (x-1{)}^2+y^2=1$
$\Rightarrow$ Centre $C_1=(1,0)$, radius $=1$
${\omega }_2:x^2+y^2+4x+8y+16=0$
$\iff (x+2{)}^2+(y+4{)}^2=4$
$\Rightarrow$ 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.
$M=\left(\frac{2(-2)+3(1)}{5},\frac{2(-4)+3(0)}{5}\right)$
$=\left(\frac{-1}{5},\frac{-8}{5}\right)$
$\Rightarrow p=-\frac{1}{5},q=-\frac{8}{5}$
$\Rightarrow \frac{q}{p}=8$