1. Find the center and radius of the given circle

The given circle is $x^2+y^2-4x+6y-12=0.$

Compare with the general form: $x^2+y^2+2gx+2fy+c=0.$

So, $2g=-4 \Rightarrow g=-2, \qquad 2f=6 \Rightarrow f=3.$

Hence the center is $(-g,-f)=(2,-3).$

Now complete the square: $x^2-4x+y^2+6y-12=0$ $(x-2)^2-4+(y+3)^2-9-12=0$ $(x-2)^2+(y+3)^2=25.$

Therefore, this circle has

• center $C=(2,-3)$
• radius $r=5$

So any diameter of this circle has length $2r=10.$

2. Interpret the condition

“One of the diameters of the given circle is a chord of another circle $S$ whose center is at $(-3,2)$.”

This means:

• Take a diameter of the first circle.
• Its endpoints lie on circle $S$.
• So that diameter is a chord of circle $S$.

A diameter of the first circle passes through its center $C=(2,-3)$.

Since this diameter is also a chord of circle $S$, the perpendicular from the center of circle $S$ to this chord bisects it.

For a fixed center of $S$, the longest possible chord through point $C$ occurs when the chord is perpendicular to the line joining the center of $S$ to $C$.

Here the chord length is fixed as $10$, and its midpoint is the center $C=(2,-3)$ of the first circle. So the distance from the center of circle $S$, say $O=(-3,2)$, to the chord is $OC.$

Compute: $OC=\sqrt{(2+3)^2+(-3-2)^2}$ $=\sqrt{5^2+(-5)^2}$ $=\sqrt{25+25}=5\sqrt{2}.$

3. Use the chord-length formula in circle $S$

If a chord of a circle of radius $R$ is at distance $d$ from the center, then its length is $2\sqrt{R^2-d^2}.$

Here,

• chord length $=10$
• distance from center of $S$ to the chord $=5\sqrt{2}$

So, $2\sqrt{R^2-(5\sqrt2)^2}=10.$

Divide by $2$: $\sqrt{R^2-50}=5.$

Square both sides: $R^2-50=25$ $R^2=75$ $R=5\sqrt3.$

4. Check options

• A: $5$ ❌
• B: $10$ ❌
• C: $5\sqrt2$ ❌
• D: $5\sqrt3$ ✅

Thus, the radius of circle $S$ is $\boxed{5\sqrt3}.$