The center of circle `x^2+y^2-6x+4y-12=0` is (a,b) then (2a + 3b) is

Options

A. 0
B. 2
C. 3
D. 5

Answer

To solve the problem, we need to find the center of the circle given by the equation \( x^2 + y^2 - 6x + 4y - 12 = 0 \) and then calculate \( 2a + 3b \) where \( (a, b) \) is the center of the circle.

### Step-by-Step Solution:

1. **Identify the Equation of the Circle**:
The given equation of the circle is:
\[ x^2 + y^2 - 6x + 4y - 12 = 0 \]

2. **Rearrange the Equation**:
We can rearrange the equation to match the standard form of a circle:
\[ x^2 - 6x + y^2 + 4y = 12 \]

3. **Complete the Square**:
We will complete the square for both \( x \) and \( y \).

- For \( x^2 - 6x \):
\[ x^2 - 6x = (x - 3)^2 - 9 \]

- For \( y^2 + 4y \):
\[ y^2 + 4y = (y + 2)^2 - 4 \]

Putting it all together, we have:
\[ (x - 3)^2 - 9 + (y + 2)^2 - 4 = 12 \]
Simplifying this gives:
\[ (x - 3)^2 + (y + 2)^2 - 13 = 12 \]
\[ (x - 3)^2 + (y + 2)^2 = 25 \]

4. **Identify the Center**:
From the standard form \( (x - h)^2 + (y - k)^2 = r^2 \), we can see that the center \( (h, k) \) of the circle is:
\[ (3, -2) \]
Therefore, \( a = 3 \) and \( b = -2 \).

5. **Calculate \( 2a + 3b \)**:
Now we need to calculate \( 2a + 3b \):
\[ 2a + 3b = 2(3) + 3(-2) \]
\[ = 6 - 6 = 0 \]

### Final Answer:
Thus, the value of \( 2a + 3b \) is \( 0 \).