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We have to find the locus when all three points are variable? In this case, the locus is the entire disc bounded by $k$ (except maybe for the actual circle; that can be reached only when the two points on it coincide). To see this: fix a point $X$ on $AB$ and a line $\ell_X$ through $X$, and make the two points $Y,Z$ on $k$ move so that $\ell_X$ is the angle bisector of $\angle YXZ$. Let $\{P,Q\}=\ell_X\cap k$. When $Y,Z\to P,Q$, we get the incenters $\to P,Q$ respectively, so, by continuity, the entire segment cut by $k$ on $\ell_X$ belongs to the locus. The only problem could arise when the incenter is $X$. This happens iff $X,Y,Z$ are collinear, so maybe we should disregard $X$, but it can be obtained as the incetner of another triangle constructed in a similar fashion by fixing a point $X'\ne X$ on $AB$, and taking $\ell_{X'}=AB$.