In $\triangle ABC$, $\angle C = 90^\circ$ and $AB = 12$. Squares $ABXY$ and $CBWZ$ are constructed outside of the triangle. The points $X$, $Y$, $Z$, and $W$ lie on a circle. What is the perimeter of the triangle?
$\textbf{(A)}\; 12+9\sqrt{3} \qquad\textbf{(B)}\; 18+6\sqrt{3} \qquad\textbf{(C)}\; 12+12\sqrt{2} \qquad\textbf{(D)}\; 30 \qquad\textbf{(E)}\; 32$
First, we should find the center and radius of this circle. We can find the center by drawing the perpendicular bisectors of $WZ$ and $XY$ and finding their intersection point. This point happens to be the midpoint of $AB$, the hypotenuse. Let this point be $M$. To find the radius, determine $MY$, where $MY^{2} = MA^2 + AY^2$, $MA = \frac{12}{2} = 6$, and $AY = AB = 12$. Thus, the radius $=r =MY = 6\sqrt5$.
Next we let $AC = b$ and $BC = a$. Consider the right triangle $ACB$ first. Using the Pythagorean theorem, we find that $a^2 + b^2 = 12^2 = 144$. 

Now, we let $E$ be the midpoint of $WZ$, and we consider right triangle $ZEM$. By the Pythagorean theorem, we have that $\left(\frac{a}{2}\right)^2 + \left(a + \frac{b}{2}\right)^2 = r^2 = 180$. Expanding this equation, we get that
\[\frac{1}{4}(a^2+b^2) + a^2 + ab = 180\]
\[\frac{144}{4} + a^2 + ab = 180\]
\[a^2 + ab = 144 = a^2 + b^2\]
\[ab = b^2\]
\[b = a\]
This means that $ABC$ is a 45-45-90 triangle, so $a = b = \frac{12}{\sqrt2} = 6\sqrt2$. Thus the perimeter is $a + b + AB = 12\sqrt2 + 12$ which is answer $\boxed{\textbf{(C)}\; 12 + 12\sqrt2}$.