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!

-
Found something. See this related question, perhaps strongly related. mathoverflow.net/questions/24307/… Note that, in the upper half plane model, if you take your points $A,B$ as being $0, \infty$ along the imaginary axis, you and they get the same answer: the locus is a slanted ray beginning at $0.$ –  Will Jagy Jul 31 '10 at 18:27
Anyway, here is a link to the first page of the article that answers the other question (constant area instead of constant opposite angle) springerlink.com/content/h031vf8c7ydfaxrf –  Will Jagy Jul 31 '10 at 21:56
The next easiest, fixing $A = i$ and $B = \infty$ along the imaginary axis, forcing a right angle at $P = x + i y$ with $x \geq 0$ gives the hyperbola $y = \sqrt{1 + x^2}.$ So the answer to your question is at least slightly different from that of question 24307. If the answer with $A = i$ and $B = \lambda i$ with finite real $\lambda > 0$ is a conic section other than a circular arc, well, I do not know the meaning of those in the upper half plane model, but I bet there are books that say. –  Will Jagy Aug 1 '10 at 1:31
Google books pulled up the following page from Richter-Gebert's recent book "Perspectives on Projective Geometry", though I can't access most of the discussion: books.google.com/… –  j.c. Jan 25 '12 at 10:33

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.