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 write the given circle in its standard form. Starting with the equation

$$x^2 + y^2 - 4x + 6y - 12 = 0,$$

we collect the $$x$$-terms and the $$y$$-terms and then complete the squares.

$$x^2 - 4x \;+\; y^2 + 6y \;=\; 12.$$

To complete the square in $$x$$ we add and subtract $$\left(\dfrac{4}{2}\right)^2 = 4$$, and to complete the square in $$y$$ we add and subtract $$\left(\dfrac{6}{2}\right)^2 = 9$$:

$$\bigl(x^2 - 4x + 4\bigr) - 4 \;+\; \bigl(y^2 + 6y + 9\bigr) - 9 \;=\; 12.$$

Simplifying, we have

$$\bigl(x - 2\bigr)^2 + \bigl(y + 3\bigr)^2 - 13 = 12,$$

so

$$\bigl(x - 2\bigr)^2 + \bigl(y + 3\bigr)^2 = 25.$$

Hence, the centre of this circle is $$C_1(2,\,-3)$$ and its radius is

$$r_1 = \sqrt{25} = 5.$$

Let the required circle $$S$$ have centre $$O(-3,\,2)$$ and radius $$r_2$$. The problem states that one of the diameters of the first circle is a chord of the circle $$S$$. Let that diameter be $$AB$$. Because $$AB$$ is a diameter of the first circle, its midpoint is exactly the centre $$C_1$$. Therefore $$C_1$$ is also the midpoint of the chord $$AB$$ of circle $$S$$.

A standard fact about chords says:
“For any chord of a circle, the line joining the centre of the circle to the midpoint of the chord is perpendicular to the chord, and if $$d$$ is the distance from the centre to the chord and $$\ell$$ is half the length of the chord, then $$r^2 = d^2 + \ell^2$$, where $$r$$ is the radius of the circle.”

We now compute each quantity for circle $$S$$.

• The distance $$d$$ from its centre $$O(-3,\,2)$$ to the midpoint $$C_1(2,\,-3)$$ is

$$d = OC_1 = \sqrt{(2 - (-3))^2 + (-3 - 2)^2} = \sqrt{5^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}.$$

• The chord $$AB$$ is a diameter of the first circle, so the full length $$AB$$ equals $$2r_1 = 2 \times 5 = 10$$, and therefore half the chord length is

$$\ell = \frac{AB}{2} = 5.$$

Applying the stated relation for chords, we substitute $$d = 5\sqrt{2}$$ and $$\ell = 5$$:

$$r_2^2 = d^2 + \ell^2 = \bigl(5\sqrt{2}\bigr)^2 + 5^2 = 50 + 25 = 75.$$

Taking the square root, we get

$$r_2 = \sqrt{75} = 5\sqrt{3}.$$

Hence, the correct answer is Option D.