If a diameter of the circle ${\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}+6\mathrm{y}-12=0$ is a chord of a circle S whose centre is at (-3,2), then the radius of S is

For the circle $x^2+y^2-4x+6y-12=0,\text{ centre }=C_1(2,-3)$

radius $r_1=\sqrt{4+9+12}=5$

Let $C_2(-3,2)$ be the centre and r, be the radius of the second circle

$\begin{array}{l}{C}_1{C}_2=\sqrt{(2+3{)}^2+(-3-2{)}^2}=5\sqrt{2}\\ \therefore r_2=\sqrt{(5\sqrt{2}{)}^2+(5{)}^2}=\sqrt{50+25}=\sqrt{75}=5\sqrt{3}.\end{array}$