Key Concepts and Formulas

• Equation of a Circle: The general form of a circle's equation is $x^2 + y^2 + 2gx + 2fy + c = 0$, where the center is $(-g, -f)$ and the radius is $\sqrt{g^2 + f^2 - c}$. The standard form is $(x-h)^2 + (y-k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.
• Distance Formula: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
• Chord Property: A line from the center of a circle perpendicular to a chord bisects the chord.

Step-by-Step Solution

Step 1: Find the center and radius of the circle $x^2 + y^2 - 4x + 6y - 12 = 0$.

We rewrite the equation in the standard form by completing the square. $(x^2 - 4x) + (y^2 + 6y) = 12$ $(x^2 - 4x + 4) + (y^2 + 6y + 9) = 12 + 4 + 9$ $(x - 2)^2 + (y + 3)^2 = 25$ $(x - 2)^2 + (y + 3)^2 = 5^2$ Thus, the center of this circle, which we'll call $C_1$, is $A(2, -3)$ and its radius $R_1$ is $5$. This step is essential to determine the properties of the circle whose diameter is a chord of the other circle.

Step 2: Visualize the problem and identify the relevant geometric relationship.

Let circle $S$ have center $O(-3, 2)$ and radius $R_S$. A diameter of circle $C_1$ is a chord of circle $S$. Since the diameter passes through the center $A(2,-3)$ of circle $C_1$, we have a chord of circle $S$ that passes through $A$. The line segment $OA$ connects the center of circle $S$ to the midpoint of the chord (which is also the center of circle $C_1$). Therefore, $OA$ is perpendicular to the chord. This creates a right triangle with hypotenuse $R_S$, one leg $OA$, and the other leg equal to the radius of $C_1$ which is 5.

Step 3: Calculate the distance between the centers of the two circles, $OA$.

Using the distance formula: $OA = \sqrt{(-3 - 2)^2 + (2 - (-3))^2}$ $OA = \sqrt{(-5)^2 + (5)^2}$ $OA = \sqrt{25 + 25}$ $OA = \sqrt{50} = 5\sqrt{2}$ This value will be used in the Pythagorean theorem in the next step.

Step 4: Apply the Pythagorean theorem to find the radius of circle $S$, $R_S$.

In the right triangle formed, $R_S^2 = OA^2 + R_1^2$. $R_S^2 = (5\sqrt{2})^2 + (5)^2$ $R_S^2 = 50 + 25$ $R_S^2 = 75$ $R_S = \sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}$

Common Mistakes & Tips

• Sign Errors: Be careful with signs when extracting the center from the general form of the circle's equation.
• Incorrect Visualization: A clear diagram is essential for understanding the geometric relationship between the two circles.
• Confusing Radii: Clearly distinguish between the radius of the first circle and the radius of the second circle.

Summary

We found the center and radius of the first circle by completing the square. We then used the fact that a diameter of the first circle is a chord of the second circle to create a right triangle. Using the distance formula, we calculated the distance between the centers of the two circles. Finally, we applied the Pythagorean theorem to find the radius of the second circle, which is $5\sqrt{3}$.

Final Answer

The final answer is \boxed{5\sqrt{3}}, which corresponds to option (D).