To find the center of the circle given by the equation \(x^2 + y^2 + 6x + 4y - 3 = 0\), we need to complete the square for both the \(x\) and \(y\) terms. Here is the step-by-step solution:
1. Rearrange the equation to group \(x\) and \(y\) terms:
[tex]\[
x^2 + 6x + y^2 + 4y - 3 = 0
\][/tex]
[tex]\[
x^2 + 6x + y^2 + 4y = 3
\][/tex]
2. Complete the square for the \(x\)-terms:
- Take the coefficient of \(x\) (which is 6), divide by 2, and square it:
[tex]\[
\left(\frac{6}{2}\right)^2 = 3^2 = 9
\][/tex]
- Add and subtract this value inside the equation:
[tex]\[
x^2 + 6x + 9 - 9
\][/tex]
- This can be written as a perfect square:
[tex]\[
(x + 3)^2 - 9
\][/tex]
3. Complete the square for the \(y\)-terms:
- Take the coefficient of \(y\) (which is 4), divide by 2, and square it:
[tex]\[
\left(\frac{4}{2}\right)^2 = 2^2 = 4
\][/tex]
- Add and subtract this value inside the equation:
[tex]\[
y^2 + 4y + 4 - 4
\][/tex]
- This can be written as a perfect square:
[tex]\[
(y + 2)^2 - 4
\][/tex]
4. Rewrite the equation including the completed squares:
[tex]\[
(x + 3)^2 - 9 + (y + 2)^2 - 4 = 3
\][/tex]
5. Simplify the equation:
[tex]\[
(x + 3)^2 + (y + 2)^2 - 13 = 3
\][/tex]
[tex]\[
(x + 3)^2 + (y + 2)^2 = 16
\][/tex]
6. Recognize the equation in standard form of a circle:
[tex]\[
(x + 3)^2 + (y + 2)^2 = 4^2
\][/tex]
From this standard form, we can identify the center and radius of the circle. The center is given by \((h, k)\) where \((x - h)\) and \((y - k)\) translate to the circle's center coordinates.
Thus, the center of the circle is \((-3, -2)\).
The correct completion of the work is:
[tex]\[
(x + 3)^2 + (y + 2)^2 = 4^2 \text{, so the center is } (-3, -2).
\][/tex]
Therefore, the correct choice is:
[tex]\((x + 3)^2 + (y + 2)^2 = 4^2 \text{, so the center is } (-3, -2).\)[/tex]
