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My equation is the following, and I would like to find which $k$ can make it a circle.


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?

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Complete the square on $x^2+4x$ and $y^2-6y$ separately.

Take all of the numbers, and the $k$, over to the right hand side.

You will have something like $(x-a)^2 + (y-b)^2 = t$, where $a$ and $b$ are numbers and $t$ is a mixture of numbers and the letter $k$.

You need $t > 0$ for a circle.

What range of values of $k$ ensures that $t>0$?

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Completing the square $$(x+2)^2+(y-3)^2=2^2+3^2-k$$

For real circle, $9+4-k\ge0$

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\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}

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