Solution:
Let $C D$ be the diameter of $k$ such that $C D \perp A B$ and let $L \in A \widehat{C} B$. Then $\angle A L B = 2 \alpha$ is constant and we have $\angle A M B = \alpha$ since $\triangle M L B$ is isosceles. Therefore $M$ belongs to an arc of the circle $k_{1}$, from which the segment $A B$ is seen by the angle $\alpha$.

Analogously, for $L \in A \widehat{D} B$ we have $\angle A L B = 180^{\circ} - 2 \alpha$, $\angle A M B = 90^{\circ} - \alpha$ and we conclude that $M$ belongs to an arc of the circle $k_{2}$, from which the segment $A B$ is seen by the angle $90^{\circ} - \alpha$. Let $t$ be the tangent line to $k$ at the point $A$ and $t \cap k_{1} = \{A, P\}$, $t \cap k_{2} = \{A, Q\}$. Let $\lambda$ be the half-plane with respect to $t$, containing $k$. Since $L$ is between $A$ and $M$ we have $M \in \lambda$. Therefore the required locus consists of the arcs of $k_{1}$ and $k_{2}$ belonging to $\lambda$.