JEE Main 2024 — Circle Question with Solution

Given,

$S_1\equiv x^2+y^2-10x+4y+13=0$ with radius $R=4$ and its one diameter is chord of other circle $S_2$ whose centre is given by intersection of $2x+3y=12 \& 3x-2y=5$ which is $\left(3,2\right)$

Now, plotting the diagram we get,

Now, from above circle the distance between the centres is given by $l=\sqrt{{\left(5-3\right)}^2+{\left(-2-2\right)}^2}=\sqrt{20}$

Hence, radius of $S_2$ will be, $r^2={\left(\sqrt{20}\right)}^2+4^2=20+16=36$

$\Rightarrow r=6$