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 will follow these steps:

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

To rewrite it in standard form, we will complete the square for \(x\) and \(y\).

1. Group the \(x\) and \(y\) terms:
\[ (x^2 - 4x) + (y^2 + 6y) = 12 \]

2. Complete the square for \(x\):
\[ x^2 - 4x = (x - 2)^2 - 4 \]

3. Complete the square for \(y\):
\[ y^2 + 6y = (y + 3)^2 - 9 \]

4. Substitute back into the equation:
\[ (x - 2)^2 - 4 + (y + 3)^2 - 9 = 12 \]
\[ (x - 2)^2 + (y + 3)^2 - 13 = 12 \]
\[ (x - 2)^2 + (y + 3)^2 = 25 \]

Now, we have the circle in standard form:
\[ (x - 2)^2 + (y + 3)^2 = 5^2 \]
This shows that the center of the circle is \(O(2, -3)\) and the radius is \(5\).

### Step 2: Use the section formula to find the diametrically opposite point
Let the point \(A\) be \( (6, 0) \) and the diametrically opposite point be \(B(x, y)\). The center \(O(2, -3)\) divides the line segment \(AB\) in the ratio \(1:1\).

Using the section formula:
\[ O = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
where \(A(6, 0)\) and \(B(x, y)\).

### Step 3: Set up equations based on the center
From the center \(O(2, -3)\):
1. For the x-coordinate:
\[ 2 = \frac{6 + x}{2} \]
Multiply both sides by \(2\):
\[ 4 = 6 + x \implies x = 4 - 6 = -2 \]

2. For the y-coordinate:
\[ -3 = \frac{0 + y}{2} \]
Multiply both sides by \(2\):
\[ -6 = y \]

### Step 4: Conclusion
Thus, the coordinates of the point diametrically opposite to \( (6, 0) \) on the circle are:
\[ B(-2, -6) \]

### Final Answer
The point diametrically opposite to the point \( (6, 0) \) on the circle is \( (-2, -6) \).

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