### Instruction: What is the circumference of the circle defined by $x^2+6x +y^2-12y-4=0$?### Response: To find the circumference of the circle, I need to know its radius.
I can use the standard form of a circle equation, $(x-h)^2+(y-k)^2=r^2$, where $(h,k)$ is the center and $r$ is the radius, to rewrite the given equation.
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 $y^2-12y$, which is the same as $(y-6)^2-36$.
So the given equation becomes $(x+3)^2-9+(y-6)^2-36-4=0$.
Simplifying, I get $(x+3)^2+(y-6)^2=49$.
Comparing this to the standard form, I can see that the center of the circle is $(-3,6)$ and the radius is $\sqrt{49}=7$.
The circumference of the circle is $2\pi r=2\pi(7)=14\pi$.