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Let $O$ be the centre of the circle $x ^2+ y ^2= r ^2$, where $r >\frac{\sqrt{5}}{2}$. Suppose $P Q$ is a chord of this circle and the equation of the line passing through $P$ and $Q$ is $2 x+4 y=5$. If the centre of the circumcircle of the triangle $O P Q$ lies on the line $x+2 y=4$, then the value of $r$ is. . . .
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Solution
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$M – I$
$OA =\frac{\sqrt{5}}{2} \quad OC =\frac{4}{\sqrt{5}}$
$CQ = OC =\frac{4}{\sqrt{5}} \text { and } CA =\frac{3}{2 \sqrt{5}}$
$\therefore \quad O Q=\sqrt{ OA ^2+ AQ ^2}=\sqrt{ OA ^2+\left( CQ ^2- CA ^2\right)}$
$\Rightarrow \quad \sqrt{\frac{5}{4}+\frac{16}{5}-\frac{9}{20}}=\sqrt{4}$
$\Rightarrow \quad 2= r$
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$P Q: h x+k y=r^2$
$\text { Given } P Q \quad 2 x+4 y=5$
$\Rightarrow \quad \frac{h}{2}=\frac{k}{4}=\frac{r^2}{5} \Rightarrow h=\frac{2 r^2}{5}\quad k=\frac{4 r^2}{5}$
$\therefore \quad C=\left(\frac{r^2}{5}, \frac{2 r^2}{5}\right)$
$\therefore \quad C \text { lies on } x+2 y=4 \quad \Rightarrow \quad \frac{r^2}{5}+2\left(\frac{2 r^2}{5}\right)=4$
$\Rightarrow \quad r^2=4 \Rightarrow r=2$
