Consider the circle defined by the equation $x^2 +6x +y^2 +8y =0$.  Find the sum of the coordinates of the center of the circle.
To find the center of the circle, I need to rewrite the equation in standard form, $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)$ is the center and $r$ is the radius. To do that, I need to complete the square for both $x$ and $y$ terms. For the $x$ terms, I have $x^2 + 6x$, which is the same as $(x+3)^2 - 9$. For the $y$ terms, I have $y^2 + 8y$, which is the same as $(y+4)^2 - 16$. So, the equation becomes $(x+3)^2 - 9 + (y+4)^2 - 16 = 0$. Adding $25$ to both sides, I get $(x+3)^2 + (y+4)^2 = 25$. This is now in standard form, so I can see that the center is $(-3,-4)$ and the radius is $5$. The sum of the coordinates of the center is $-3 + (-4) = -7$. 