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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 $M{Y}^2=M{A}^2+A{Y}^2$, $MA=\frac{12}{2}=6$, and $AY=AB=12$. Thus, the radius $=r=MY=6\sqrt{5}$. 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 $a=b=\frac{12}{\sqrt{2}}=6\sqrt{2}$. Thus the perimeter is $a+b+AB=12\sqrt{2}+12$ which is answer $\boxed{12+12\sqrt{2}}$.