Question

Let $O$ be the centre of the circle $x^2+y^2=r^2,$ where $r>\frac{\sqrt{5}}{2}.$ Suppose $PQ$ is a chord of this circle and the equation of the line passing through $P$ and $Q$ is $2x+4y=5.$ If the centre of the circumcircle of the triangle $OPQ$ lies on the line $x+2y=4,$ then the value of $r$ is

Answer 1

$S_1:x^2+y^2=r^2$ where $r>\frac{\sqrt{5}}{2}$ and centre $C_1=(0,0)$
Now, let $S_2:x^2+y^2+ax+by=0$
$\Rightarrow$ Centre $C_2=(\frac{-a}{2},\frac{-b}{2})$
Radical axis of $S_1=0$ and $S_2=0$ is $PQ$
$PQ:S_1-S_2=0$
$PQ:ax+by+r^2=0\cdots \left(1\right)$
Given $PQ=2x+4y-5=0\cdots \left(2\right)$
On comparing equation $\left(1\right)$ and $\left(2\right)$
$\frac{a}{2}=\frac{b}{4}=\frac{r^2}{-5}\Rightarrow -\frac{a}{2}=\frac{r^2}{5},-\frac{b}{2}=\frac{2r^2}{5}\cdots \left(3\right)$

Also, centre of $S_2$ lies on $x+2y=4$
$\Rightarrow \frac{r^2}{5}+\frac{4r^2}{5}=4\Rightarrow r=2$