If one of the diameter 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 (2,-3), the radius of S is

Options

A. `sqrt(41)` units
B. `3sqrt(5)` units
C. `5sqrt(2)` units
D. `2sqrt(5)` units

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Rewrite the equation of the circle in standard form
The given equation of the circle is:
\[ x^2 + y^2 + 4x + 6y - 12 = 0 \]

We can rearrange this equation to find the center and radius of the circle. We will complete the square for both \(x\) and \(y\).

### Step 2: Completing the square for \(x\) and \(y\)
1. For \(x\):
\[ x^2 + 4x \quad \text{can be written as} \quad (x + 2)^2 - 4 \]

2. For \(y\):
\[ y^2 + 6y \quad \text{can be written as} \quad (y + 3)^2 - 9 \]

### Step 3: Substitute back into the equation
Substituting these completed squares back into the equation gives:
\[ (x + 2)^2 - 4 + (y + 3)^2 - 9 - 12 = 0 \]
This simplifies to:
\[ (x + 2)^2 + (y + 3)^2 - 25 = 0 \]
Thus, we can rewrite it as:
\[ (x + 2)^2 + (y + 3)^2 = 25 \]

### Step 4: Identify the center and radius of the first circle
From the standard form \((x - h)^2 + (y - k)^2 = r^2\), we can identify:
- Center: \((-2, -3)\)
- Radius: \(r = \sqrt{25} = 5\)

### Step 5: Determine the diameter
The diameter of the first circle is \(AB\) which is a chord of the second circle \(S\) whose center is given as \((2, -3)\).

### Step 6: Find the distance between the centers of the two circles
Using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the centers:
\[ d = \sqrt{(2 - (-2))^2 + (-3 - (-3))^2} = \sqrt{(2 + 2)^2 + (0)^2} = \sqrt{4^2} = 4 \]

### Step 7: Use the Pythagorean theorem to find the radius of circle \(S\)
Let \(R\) be the radius of circle \(S\). The relationship between the radius of circle \(S\), the radius of the first circle (5), and the distance between the centers (4) can be expressed as:
\[ R^2 = 5^2 + 4^2 \]
Calculating this gives:
\[ R^2 = 25 + 16 = 41 \]
Thus, the radius \(R\) is:
\[ R = \sqrt{41} \]

### Final Answer
The radius of circle \(S\) is \(\sqrt{41}\).

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