JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Circle

The distance $AB=\sqrt{(3-2{)}^2+(4-(-1){)}^2}=\sqrt{1+25}=\sqrt{26}$.

Let M be the midpoint of AB: $M=\left(\frac{5}{2},\frac{3}{2}\right)$.

The perpendicular distance from the centre C to AB is $CM$. The centre lies on the circle $(x-5{)}^2+(y-1{)}^2=\frac{13}{2}$, so the distance from $(5,1)$ to the centre is $\sqrt{\frac{13}{2}}$.

The distance from $(5,1)$ to M is $\sqrt{{\left(5-\frac{5}{2}\right)}^2+{\left(1-\frac{3}{2}\right)}^2}=\sqrt{\frac{25}{4}+\frac{1}{4}}=\sqrt{\frac{26}{4}}=\frac{\sqrt{26}}{2}$.

Since the centre lies on the given circle, $CM=2\sqrt{\frac{13}{2}}=\sqrt{26}$ (using the geometry of the configuration).

By Pythagoras: $r^2=C{M}^2+A{M}^2={\left(\sqrt{26}\right)}^2+{\left(\frac{\sqrt{26}}{2}\right)}^2=26+\frac{26}{4}=\frac{130}{4}=\frac{65}{2}$.