Let $A \subset \mathbb{R}^2$ be a set of points given by equation $f(x,y) = 0$, then for any invertible linear transformation $T : \mathbb{R}^2 \to \mathbb{R}^2$ set $T(A)$ is given by $(f \circ T^{-1})(x,y) = 0$.
This is because $A = \{(x,y) \mid f(x,y) = 0\}$, and
\begin{align}
T(A) &= \Big\{(x',y')\ \Big|\ (x',y') = T(x,y) \land f(x,y) = 0\Big\} \\
&= \Big\{(x',y')\ \Big|\ T^{-1}(x',y') = (x,y) \land f(x,y) = 0\Big\} \\
&= \Big\{(x',y')\ \Big|\ f(T^{-1}(x',y')) = 0\Big\} \\
&= \Big\{(x',y')\ \Big|\ (f \circ T^{-1})(x',y') = 0\Big\}.
\end{align}
For example, if $T = \left[\begin{array}{rr}0&-1\\-1&0\end{array}\right]$, that is the symmetry along $y = -x$, then it happens that $T^{-1} = T$ (this is true for all symmetries) and appropriate equation would look like $$f(-y,-x) = 0.$$
In your case $f(x,y) = (x+3)^2 + (y-3)^2 -9$, so the symmetrical figure could be described by $$(-y+3)^2+(-x-3)^2 = 9.$$
I hope this helps ;-)