Vectorial solution is a cinch. Take $M_{\omega}$ on $[OX$ and $N_{\omega}$ on $[OY$ such that $|OM_{\omega}| = |ON_{\omega| =}$ half that constant. The locus is the segment $[M_{\omega}N_{\omega}]$.

Let $M'$ and $N'$ be sample points such that $OM'=ON' = \dfrac k2$. Since $k$ is constant, $\angle M'ON'$ is constant, so $[M'N']$ is constant. Draw parallel through $N$ to $OM$. Suppose this parallel meet $M'N'$ at $Q$. Let $MN$ meet $M'N'$ at $P$. Since $OM+ON=OM'+ON'=2OM'=2ON'$, $MM'=MM'$. Since $\angle OM'N' = \angle NQN'$, $NQ=NN'=MM'$. So $MP=PN$. This means midpoint of the segments $MN$ are always on $M'N'$. Now, let's show the inverse. We will show for any point $P$ on $[M'N']$, there are $M$ and $N$ such that $OM+ON=k$. Let the reflection of $O$ over $P$ be $O'$. Let parallel though $O'$ to $ON'$ meet $OM'$ at $M$. Let parallel though $O'$ to $OM'$ meet $ON'$ at $N$. Since $ONO'M$ is parallelogram, $P$ is the midpoint of $MN$. Now we will show that $OM+ON=k$. Let $O'N$ meet $M'N'$ at $Q$. Since $OM'N'$ is isosceles, so $NQN'$ is isosceles. So $NQ=N'N$. Also $MM'=NQ$, so $OM+ON=OM'+ON'=k$.

This exercise is a part of next story: Let $\vec{v}$ be fixed on line $OX$ with lenght $|OM|+|ON|$. Then we get $N$ with translation by $\vec{v}$ from $M$ to $M'$ and then reflection of $M'$ on external angle bisector of $\angle XOY$. So $N$ is picture of $M$ under indirect isometry so the midpoint $P$ of segment $MN$ describe, by Hjelmslev's theorem, line. In particular case, it is easy to see that this is segment $M_{\omega}N_{\omega}$ since: if $N=O$ then $P = M_{\omega}$ and if $M=O$ then $P = N_{\omega}$. See also this.