Let $CA,CB$ be tangent to the smaller circle at $X,Y$. By a well known lemma, since the smaller circle is the $C$- mixtillinear incircle of $ABC$, its incenter is the midpoint of $XY$, or the inverse of $C$ about the smaller circle. But then the locus of the incenters is simply the inverse of the bigger circle about the smaller one. EDIT: Misread and assumed that the two circles were tangent.

Assume that the smaller circle is $(I;r)$ and the external circle is $(O;R)$, the incenter of $\triangle CAB$ be $J$. We shall prove that $IJ.IC=const$. Let $CI \cap (O)=P$, hence $PJ=PA=PB$. We assume that $J$ lies between $I$ and $P$ (the other case can be proved similarly). Let $CA$ meets $(I)$ at $Q$. We have $IP.IC=R^2-OI^2$ (1) $\frac{IJ}{IP}=1-\frac{PI}{PJ}=1-\frac{PI}{PA}$. (2) On the other hand, $\frac{IP}{PA}=\frac{IP}{2R.sinACP}=\frac{R^2-OI^2}{2R.IC.sinACP}=\frac{R^2-OI^2}{2R.IQ}=\frac{R^2-OI^2}{2Rr}=const$. (3) From (1), (2) and (3), we can calculate the value $IJ.IC=\frac{(2Rr+OI^2-R^2)(R^2-OI^2)}{2Rr}=k$, so $J$ lies on the circle which is the image of $(O)$ through the inversion center $I$, radius $k$.

Label $(I,r)$ and $(O,R)$ the given circles, this latter being the external one. $J$ is the incenter of $\triangle ABC$ and $CJ$ cuts $(O)$ again at $D.$ Keeping in mind that $DA=DB=DJ$ (well-known), we get $IC \cdot IJ=IC \cdot (ID-JD)=IC \cdot ID-IC \cdot DA=$ $=IC \cdot ID-\frac{r}{\sin \widehat{ACD}} \cdot 2R \cdot \sin \widehat{ACD}=IC \cdot ID-2R \cdot r=\text{const}=-\varrho^2.$ Thus, locus of $J$ is the inverse circle of $(O)$ under inversion with center $I$ and power $-\varrho^2.$