The point diametrically opposite to the point (6, 0) on the circle `x^(2) +y^(2)-4x+6y-12=0` is :

Options

A. (2, 6)
B. (-2, 6)
C. (2, -6)
D. (-2, -6)

Explanation

To find the point diametrically opposite to the point (6, 0) on the given circle, we can follow these steps:

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

We can rewrite this in standard form by completing the square.

### Step 2: Complete the Square
1. For the \(x\) terms:
\[ x^2 - 4x \]
Completing the square:
\[ (x - 2)^2 - 4 \]

2. For the \(y\) terms:
\[ y^2 + 6y \]
Completing the square:
\[ (y + 3)^2 - 9 \]

Now substituting these back into the equation:
\[ (x - 2)^2 - 4 + (y + 3)^2 - 9 - 12 = 0 \]
\[ (x - 2)^2 + (y + 3)^2 - 25 = 0 \]
\[ (x - 2)^2 + (y + 3)^2 = 25 \]

### Step 3: Identify the Center and Radius
From the equation \((x - 2)^2 + (y + 3)^2 = 25\), we can identify:
- Center of the circle: \(C(2, -3)\)
- Radius: \(r = 5\) (since \(25 = 5^2\))

### Step 4: Use the Midpoint Formula
Let the point diametrically opposite to (6, 0) be \(P(x, y)\). The center \(C\) is the midpoint of the diameter formed by points (6, 0) and \(P(x, y)\).

Using the midpoint formula:
\[ C = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]
Substituting \(C(2, -3)\) and \(P(6, 0)\):
\[ \left(\frac{6 + x}{2}, \frac{0 + y}{2}\right) = (2, -3) \]

### Step 5: Set Up the Equations
From the midpoint formula, we can set up the following equations:
1. \(\frac{6 + x}{2} = 2\)
2. \(\frac{0 + y}{2} = -3\)

### Step 6: Solve for \(x\) and \(y\)
1. For the first equation:
\[ 6 + x = 4 \implies x = 4 - 6 = -2 \]

2. For the second equation:
\[ y = -6 \]

### Step 7: Conclusion
Thus, the point diametrically opposite to (6, 0) on the circle is:
\[ \boxed{(-2, -6)} \]