Problem
Mrs. Culland is finding the center of a circle whose equation is $x^{2}+y^{2}+6 x+4 y-3=0$ by completing the square. Her work is shown. \[ \begin{array}{l} x^{2}+y^{2}+6 x+4 y-3=0 \\ x^{2}+6 x+y^{2}+4 y-3=0 \\ \left(x^{2}+6 x\right)+\left(y^{2}+4 y\right)=3 \\ \left(x^{2}+6 x+9\right)+\left(y^{2}+4 y+4\right)=3+\underline{9}+4 \end{array} \] Which completes the work correctly? $(x-3)^{2}+(y-2)^{2}=4^{2}$, so the center is $(3,2)$. $(x+3)^{2}+(y+2)^{2}=4^{2}$, so the center is $(3,2)$. $(x-3)^{2}+(y-2)^{2}=4^{2}$, so the center is $(-3,-2)$. $(x+3)^{2}+(y+2)^{2}=4^{2}$, so the center is $(-3,-2)$.
Solution
Step 1 ：Given the equation of the circle as $x^{2}+y^{2}+6 x+4 y-3=0$
Step 2 ：Rearrange the equation as $x^{2}+6 x+y^{2}+4 y-3=0$
Step 3 ：Group the x terms and y terms together to get $(x^{2}+6 x)+(y^{2}+4 y)=3$
Step 4 ：Complete the square for the x terms and y terms to get $(x^{2}+6 x+9)+(y^{2}+4 y+4)=3+9+4$
Step 5 ：Simplify the equation to get $(x+3)^{2}+(y+2)^{2}=4^{2}$
Step 6 ：From the equation, we can see that the center of the circle is at the point (-3, -2)
Step 7 ：\(\boxed{\text{The correct completion of Mrs. Culland's work is }"(x+3)^{2}+(y+2)^{2}=4^{2}", \text{ so the center is } (-3,-2)}\)
