It is known that if the angle subtended from two foci $ \text{ in the sketch} (\pm 3, -4) $ at origin and at the arc are $ (\gamma, \gamma/2 ) $ respectively then the curve is a circular arc.
What are curves if the angles subtended on them are $ \gamma/4, \gamma/6 $ etc. instead?
Are these curves known?
Plotted here are boundary angle fractions found by dot product of the enclosing vectors.
They're circular arcs as well: referencing your diagram, if you want the angle to be $k\gamma$ continue vertically from your point $O$ until you hit the angle $\widehat{O'}=2k\gamma$ at point $O'$. Then, your locus is an arc of the circle with center $O'$ and passing trhough your two original "foci" (and symmetrically for the part in the lower half-space).
Addendum: Removing the unnecessary dependence on $\gamma$, more generally, given a segment $AB$, the locus of the points that see the segment by an angle $0<\alpha<\pi$ may be devised as such:
call $U_1$ and $U_2$ the two points such that $\triangle AU_iB$ is an isosceles triangle with basis $AB$ and $\angle AU_iB=\frac\pi2-\alpha$.
your locus is the set of points $P$ satisfying the identity $\min\left(\overline{PU_1},\overline{PU_2}\right)=\overline{U_1A}$, which is the union of two circles, minus the arcs lying inside either of the disks.
This assumes the limit-case convention that $A$ and $B$ should see $AB$ by any angle.
Notice that the set $\{U_1,U_2\}$ is completely determined by the fact that $\overline{U_iA}=\overline{AB}\csc \alpha$. Namely, if we set cartesian coordinates such that $A=\left(-\frac12,0\right)$ and $B=\left(\frac12,0\right)$, then we may take $U_i=\left(0,\frac{(-1)^i\cot\alpha}{2}\right)$ and the locus is described by the equation $$\min\left(\sqrt{x^2+\left(y-\frac{\cot \alpha}2\right)^2},\sqrt{x^2+\left(y+\frac{\cot \alpha}2\right)^2}\right)=\csc \alpha$$ or, after some algebra, $$x^2+\left(\lvert y\rvert-\frac12\lvert \cot\alpha\rvert\right)^2=\csc^2\alpha$$