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 : (1) `5sqrt(2)` (2) `5sqrt(3)` (3) 5 (4) 10

Options

A. `5 sqrt(3)`
B. 5
C. 10
D. `5 sqrt(2)`

Explanation

To solve the problem, we need to find the radius of the circle \( S \) whose center is at \((-3, 2)\) and has a chord that is a diameter of another circle given by the equation \( x^2 + y^2 - 4x + 6y - 12 = 0 \).

### Step 1: Rewrite the equation of the circle in standard form
The given equation is:
\[ x^2 + y^2 - 4x + 6y - 12 = 0 \]
We can rearrange it to find the center and radius. First, we complete the square for \( x \) and \( y \).

For \( x \):
\[ x^2 - 4x \rightarrow (x - 2)^2 - 4 \]

For \( y \):
\[ y^2 + 6y \rightarrow (y + 3)^2 - 9 \]

Substituting back, we have:
\[ (x - 2)^2 - 4 + (y + 3)^2 - 9 - 12 = 0 \]
\[ (x - 2)^2 + (y + 3)^2 - 25 = 0 \]
\[ (x - 2)^2 + (y + 3)^2 = 25 \]

### Step 2: Identify the center and radius of the first circle
From the standard form \( (x - h)^2 + (y - k)^2 = r^2 \):
- Center \( C_1 = (2, -3) \)
- Radius \( R_1 = 5 \) (since \( r^2 = 25 \))

### Step 3: Find the distance between the centers of the two circles
The center of circle \( S \) is given as \( C = (-3, 2) \). We need to find the distance \( CC_1 \) between \( C \) and \( C_1 \).

Using the distance formula:
\[ CC_1 = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ CC_1 = \sqrt{((-3) - 2)^2 + (2 - (-3))^2} \]
\[ = \sqrt{(-5)^2 + (5)^2} \]
\[ = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \]

### Step 4: Apply the Pythagorean theorem to find the radius of circle \( S \)
In triangle \( AC C_1 \), where \( A \) and \( B \) are endpoints of the diameter of the first circle, we can apply the Pythagorean theorem:
\[ AC^2 + C_1C^2 = R_1^2 \]
Let \( AC \) be the radius of circle \( S \) (which we need to find). Thus,
\[ AC^2 + (5\sqrt{2})^2 = 5^2 \]
\[ AC^2 + 50 = 25 \]
\[ AC^2 = 25 - 50 = -25 \]
This indicates that we made a mistake in the application of the theorem; we should instead consider:
\[ AC^2 = C_1C^2 + R_1^2 \]
Thus:
\[ AC^2 = (5\sqrt{2})^2 + 5^2 \]
\[ = 50 + 25 = 75 \]
\[ AC = \sqrt{75} = 5\sqrt{3} \]

### Conclusion
The radius of circle \( S \) is \( 5\sqrt{3} \).