Let $M$ be the midpoint of arc $AB$ not containing $C$ and $D$. Observe that $\angle{AI_1B}=90+\frac{1}{2}\angle{ACB}=90+\frac{1}{2}\angle{ADB}=\angle{AI_2B}$. Thus $AI_1I_2B$ is cyclic and $M$ lies on $CI_1$ and $DI_2.$ Since $CD$ is constant, $I_1I_2$ is also constant. By the incenter/excenter lemma, we have $MA=MI_1=MI_2=MB$. Let $N$ be the midpoint of $I_1I_2$, then $MN\perp I_1I_2$. Let $\gamma$ be the circle with center $M$ and radius $MN$, then note that $I_1I_2$ is tangent to $\gamma$. As $C$ varies over arc $AB$, the length $MN$ remains constant. Thus $\gamma$ is a fixed circle which gives us our desired result.