Equation with k describing a circle My equation is the following, and I would like to find which $k$ can make it a circle.
$$x^2+y^2+4x-6y+k=0$$
My naive approach is to have $k$ to be $-4x+6y+c$ where c is any number, so that I can have any circle that is in 0. However k is a parameter and I can't really figure that out if I am missing something. Any advice?

\begin{align}
x^2+y^2+4x-6y+k&=0\\
x^2+4x+y^2-6y&=-k\\
x^2+4x+4+y^2-6y+9&=-k+4+9\\
(x+2)^2+(y-3)^2&=13-k
\end{align}
Compare with equation of the circle where its center on $(a,b)$ and radius $r$.
\begin{align}
(x-a)^2+(y-b)^2=r^2
\end{align}
We get $r^2=13-k$. In order to make a circle, then $r>0$.
\begin{align}
r&>0\\
\sqrt{13-k}&>0\\
13-k&>0\\
k&<13.
\end{align}
