Circle $k$ and its diameter $AB$ are given. Find the locus of the centers of circles inscribed in the triangles having one vertex on $AB$ and two other vertices on $k.$
1. **Define the problem and setup:**
   - We are given a circle \( k \) with diameter \( AB \).
   - We need to find the locus of the centers of circles inscribed in triangles that have one vertex on \( AB \) and the other two vertices on \( k \).

2. **Consider a point \( X \) on \( AB \):**
   - Let \( X \) be a variable point on \( AB \).
   - Draw a line \( \ell_X \) through \( X \). This line will act as the angle bisector of \( \angle YXZ \) where \( Y \) and \( Z \) are points on \( k \).

3. **Intersection of \( \ell_X \) with \( k \):**
   - Let \( \{P, Q\} = \ell_X \cap k \).
   - As \( Y \) and \( Z \) move along \( k \) such that \( \ell_X \) remains the angle bisector, the incenter of \( \triangle XYZ \) will lie on the segment of \( \ell_X \) that is inside \( k \).

4. **Continuity argument:**
   - When \( Y \) and \( Z \) approach \( P \) and \( Q \) respectively, the incenter of \( \triangle XYZ \) approaches \( P \) and \( Q \).
   - By continuity, the entire segment of \( \ell_X \) inside \( k \) belongs to the locus of the incenters.

5. **Special case when the incenter is \( X \):**
   - The incenter is \( X \) if and only if \( X, Y, Z \) are collinear.
   - This situation can be disregarded as \( X \) can be obtained as the incenter of another triangle by fixing a different point \( X' \neq X \) on \( AB \) and taking \( \ell_{X'} = AB \).

6. **Conclusion:**
   - The locus of the centers of the inscribed circles is the entire disc bounded by \( k \), excluding the actual circle \( k \) itself.

The final answer is the entire disc bounded by \( \boxed{ k } \), excluding the actual circle \( k \) itself.