JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Circle

The chord $PQ$ has equation $2x+4y=5$. The perpendicular from $O$ to $PQ$ has slope $2$ (negative reciprocal of $-1/2$), so its equation is $y=2x$. The circumcentre $C$ of triangle $OPQ$ lies on the perpendicular bisector of $PQ$, which passes through the midpoint of $PQ$. For a chord $hx+ky=r^2$ of $x^2+y^2=r^2$, the circumcentre is at $(h,k)$. Comparing with $2x+4y=5$, we get $\frac{h}{2}=\frac{k}{4}=\frac{r^2}{5}$, so $h=\frac{2r^2}{5}$ and $k=\frac{4r^2}{5}$. Thus $C=\left(\frac{r^2}{5},\frac{2r^2}{5}\right)$. Since $C$ lies on $x+2y=4$: $\frac{r^2}{5}+2\left(\frac{2r^2}{5}\right)=4$, giving $r^2=4$, so $r=2$.