imc

The center of the circle lies on the intersection between the perpendicular bisectors of chords $ZW$ and $YX$. Therefore we know the center of the circle must also be the midpoint of the hypotenuse. Let this point be $O$. Draw perpendiculars to $ZW$ and $YX$ from $O$, and connect $OZ$ and $OY$. $O{Y}^2=6^2+12^2=180$. Let $AC=a$ and $BC=b$. Then ${\left(\frac{a}{2}\right)}^2+{\left(a+\frac{b}{2}\right)}^2=O{Z}^2=O{Y}^2=180$. Simplifying this gives $\frac{a^2}{4}+\frac{b^2}{4}+a^2+ab=180$. But by Pythagorean Theorem on $△ABC$, we know $a^2+b^2=144$, because $AB=12$. Thus $\frac{a^2}{4}+\frac{b^2}{4}=\frac{144}{4}=36$. So our equation simplifies further to $a^2+ab=144$. However $a^2+b^2=144$, so $a^2+ab=a^2+b^2$, which means $ab=b^2$, or $a=b$. Aha! This means $△ABC$ is just an isosceles right triangle, so $AC=BC=\frac{12}{\sqrt{2}}=6\sqrt{2}$, and thus the perimeter is $\boxed{12+12\sqrt{2}}$.