JEE Main 2022 — Circle Question with Solution

Equation of line joining $AB$ is 

$y+1=5\left(x-2\right)$

i.e. $5x-y-11=0$ 

Also $AB=\sqrt{{\left(3-2\right)}^2+{\left(4+1\right)}^2}=\sqrt{26}$

$r$ is the radius of the circle $C$

For the circle ${\left(x-5\right)}^2+{\left(y-1\right)}^2=\frac{13}{2}$

centre is $\left(5,1\right)$ and radius $R=\sqrt{\frac{13}{2}}$

Equation of perpendicular bisector of $AB$ will be $y-1=-\frac{1}{5}\left(x-5\right)\Rightarrow x+5y-10=0$

Solving this equation of line with ${\left(x-5\right)}^2+{\left(y-1\right)}^2=\frac{13}{2}$, we get 

$M$ as $\left(\frac{5}{2},\frac{3}{2}\right)$ 

So $M$ lies on the line joining $AB$.

Now $r^2=C{M}^2+A{M}^2$

$={\left(2\times \sqrt{\frac{13}{2}}\right)}^2+{\left(\sqrt{\frac{13}{2}}\right)}^2$

$r^2=\frac{65}{2}$