Animated Solution for Mathematics - Circles: 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:

First, we must understand our primary subject, circle $C$. The equation $x^2+y^2-4x+6y-12=0$ is in the general form $x^2+y^2+2gx+2fy+c=0$.

By comparing the coefficients, we find $2g=-4$ and $2f=6$, which gives us $g=-2$ and $f=3$. The center $C_1$ is $(-g,-f)$, so $C_1=(2,-3)$.

Now, for the radius $r$, we use the formula $r=\sqrt{g^2+f^2-c}$. Substituting our values, we get:

We have successfully decoded circle $C$: it is centered at $(2,-3)$ with a radius of $5$.

The problem states that a diameter of circle $C$ is a chord of circle $S$. This is the "Aha!" moment. A diameter of $C$ must pass through its center $C_1(2,-3)$.

Therefore, the chord of circle $S$ passes through $C_1$, and $C_1$ is the midpoint of this chord. This is a classic JEE geometry setup.

We know that the line segment joining the center of a circle to the midpoint of a chord is always perpendicular to that chord. Thus, the line segment $O{C}_1$ is perpendicular to the chord.

Now, visualize a right-angled triangle formed by the center of circle $S$ (point $O$), the center of circle $C$ (point $C_1$), and one endpoint of the chord (let us call it $A$). The distance $O{C}_1$ is the base, the segment $C_{1}A$ is the perpendicular, and the segment $OA$ is the hypotenuse.

The length $C_{1}A$ is simply the radius of circle $C$, which is $5$. The length $OA$ is the radius $R$ of circle $S$, which we are trying to find.

First, let us calculate the distance $d$ between the centers $O(-3,2)$ and $C_1(2,-3)$ using the distance formula:

With our triangle $△O{C}_{1}A$ fully defined, we apply the Pythagorean theorem: $O{A}^2=O{C}_1^2+C_1{A}^2$. Substituting our known values, we get:

This simplifies to $R^2=50+25=75$. Finally, taking the square root, we find:

The radius of circle $S$ is $5\sqrt{3}$. It is elegant, it is precise, and it is the result of understanding the geometric soul of the problem.