Poincaré disk model: is this locus a known curve? Please, consider a line segment $AB$ in the Poincaré disk model. Now, consider the set $S$ of all point $P$ in the disk such that the angle $\angle APB$ is constant. 
Question: is $S$ a known curve?
Thanks!
 A: In the Klein model, one may see that this is also a circle. Consider a line segment with one point on the center of the disk. One side of the triangle goes through the center. Then orthogonal lines to a line through the center are also orthogonal in the hyperbolic metric, e.g. since they are preserved by reflection. So one sees that a circle is traced out which goes through the origin. If you'd rather center the curve at the origin, then it will be an ellipse, since hyperbolic isometries of the Klein model are projective transformations. 
To convert to the Poincare model, take a hemisphere sitting over the disk, and project vertically. The projection of the circle is given by the intersection of a cylinder over the circle with the upper hemisphere. This upper hemisphere is conformally equivalent to the Poincare model, e.g. by inversion through a sphere centered at the south pole of the lower hemisphere. I haven't computed the curve this traces out though. 
A: I imagine the answer to this problem is known, but I don't have a reference for it; if someone has one, I would be interested to see it. Here is my calculation in the upper half-plane.
Following on from Will Jagy's comment, for a fixed angle $\theta$ at $P = x + iy$ and with the other two vertices in the upper half-plane placed at $i$ and $\infty$, the locus $S$ for the vertex $P$ is
$$y^2 = 1 + x^2 - 2xy\cot{\theta}.$$
We can use this to extend to the case where the two fixed vertices are at $i$ and $e^h i$, where $h$ is the distance between the two points $A$ and $B$. Suppose $P$ makes the fixed angle $\theta$ with these two vertices, and denote by $\psi$ the angle at $P$ of the ideal triangle with vertices at $e^h i$ and $\infty$. We then have the two equations
$$y^2 = e^{2h} + x^2 - 2xy\cot{\psi}$$
$$y^2 = 1 + x^2 - 2xy\cot{(\psi + \theta)}.$$
Using trig identities to remove $\psi$, from this we get
$$y^4 - (e^{2h} + 1)y^2 + x^4 + (e^{2h} + 1)x^2 + 2xy(xy + \cot{\theta} - e^{2h}\cot{\theta}) = e^{2h}.$$
I am not personally aware of any particular significance of this locus, though I'd be interested to hear if there is one.
