Thank you for reporting, we will resolve it shortly
Q.
If one of the diameters of the circle, given by the equation,$ x^2 + y^2 - 4x + 6y - 12 = 0$, is a chord of a circle $S$ , whose centre is at $(-3, 2)$, then the radius of $S$ is :
$x^{2}+y^{2}-4 x+6 y-12=0$
Centre $(2,-3)$
Radius $\sqrt{4+9+12}=5$
Distance $b / w$ two centres $c_{1}(2,-3)$ and $c_{2}(-3,2)$
$d=\sqrt{(2+3)^{2}+(-3-2)^{2}}=\sqrt{50}$
Radius of $( S )=\sqrt{5^{2}+(\sqrt{50})^{2}}=\sqrt{75}=5 \sqrt{3}$