What is the center of ellipse given by $||Ax-x_c||_2\leq 1$? (Cartesian plane, $x\in\mathbb{R^2}$)
I thought that to find the center I need to equate the norm to zero and then we have $x=A^{-1}x_c$ (assuming that $A$ is invertible). However, when I plot the ellipse and point $A^{-1}x_c$, it doesn't look like a center.
Why not? Let $$E=\{x:\|Ax-x_c\|_2=1\}$$ be the (boundary of the) elipse. Then $c$ is the center of $E$ if (and only if) $$x_1\in E\implies x_2=c-(x_1-c)=2c-x_1\in E,$$ that is, the ellipse is symmetric w.r.t. the point $c$.
Let's set $c:=A^{-1}x_c$ and check if we get the property satisfied. If $x_1\in E$, that is, $\|Ax_1-x_c\|_2=1$, then clearly $$\begin{split} \|Ax_2-x_c\|_2&=\|A(2c-x_1)-x_c\|_2=\|A(2A^{-1}x_c-x_1)-x_c\|_2\\&=\|x_c-Ax_1\|_2=\|Ax_1-x_c\|_2=1,\end{split}$$ so $x_2\in E$. Therefore, $A^{-1}x_c$ is indeed the center of $E$.
You should check again your plotting code.