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(D) The given circle is $x^2 + y^2 - 4x + 6y - 12 = 0$. Its center $O$ is $(2, -3)$ and its radius $r$ is $\sqrt{2^2 + (-3)^2 - (-12)} = \sqrt{4 + 9 +

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$5\sqrt{3}$

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(D) The given circle is $x^2 + y^2 - 4x + 6y - 12 = 0$. Its center $O$ is $(2, -3)$ and its radius $r$ is $\sqrt{2^2 + (-3)^2 - (-12)} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5$.Let $A(-3, 2)$ be the center of circle $S$. $A$ diameter of the given circle is a chord of circle $S$. Let this chord be $BC$. Since $BC$ is a diameter of the first circle,it passes through the center $O(2, -3)$.In $\Delta ABC$,$A$ is the center of circle $S$,so $AB$ and $AC$ are radii of $S$. $O$ is the midpoint of chord $BC$,so $AO \perp BC$.The distance $AO$ is the distance between $(-3, 2)$ and $(2, -3)$:$AO = \sqrt{(2 - (-3))^2 + (-3 - 2)^2} = \sqrt{5^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$.In the right-angled triangle $\Delta AOB$,the radius $R$ of circle $S$ is the hypotenuse $AB$:$R = \sqrt{AO^2 + OB^2} = \sqrt{(5\sqrt{2})^2 + 5^2} = \sqrt{50 + 25} = \sqrt{75} = 5\sqrt{3}$.

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 center is at $(-3, 2)$,then the radius of $S$ is:

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