Mapping a circle to curved space I'm not sure how to state my question in proper mathematical terms, so I'll try to explain it here using some picture.
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On the left you see a circle defined by its center point c and its radius r. Also we have a point p which is defined as being the point on the circle alpha radians from the positive x axis. So far this should be pretty standard.
What I want to do now, is to map the left circle onto some curved space as illustrated on the right. In this example the curvature is set to k=1. The circle's center in curved space is set where the tangent lines (dashed blue) meet. The (now curved) arc should stay 2r which results in a new r'. Also there should be some mapping from alpha to alpha', because in the lower part of the circles the angles get compressed while they get stretched in the upper part.
What would be the mapping functions (c->c'; r->r'; alpha->alpha') given the constraints for the new circle ?
Sorry for not being able to describe this problem more rigorously in mathematical terms, but there should be a sensible mapping like this (and it most likely has a name as well ;) ).
 A: Your diagrams and description are consistent with applying a Möbius transformation to a circle so that two points on a diameter are fixed. The circle sweeps out a family of circles of Apollonius.

This transformation is most easily described using complex numbers: If $p$ is a real number, and if $z = x + iy$ is complex, then the mapping
$$
f(z) = \frac{z + ip}{ipz + 1}
= \frac{x + i(y + p)}{1 - py + ipx}
= \frac{[x + i(y + p)][1 - py - ipx]}{(1 - py)^{2} + (px)^{2}}
$$
sends the unit circle ($z = e^{i\alpha}$, or $x = \cos\alpha$, $y = \sin\alpha$) to a circle passing through $\pm1$. (To handle a circle of radius $r$, replace $z$ by $z/r$.) The parameter $p$ controls the "amount of distortion". When $p = \pm1$, the distorted circle acquires infinite radius, and becomes the $x$-axis.
The bottom and top points of the distorted circle are
$$
f(-i) = i\frac{p - 1}{p + 1},\qquad
f(i) = i\frac{1 + p}{1 - p}.
$$
The center of the distorted circle is the midpoint, $\frac{1}{2}[f(i) + f(-i)]$; the radius of the distorted circle is half the magnitude of the difference, $\frac{1}{2}|f(i) - f(-i)|$. Thus
$$
\text{Center} = c' = i\frac{2p}{1 - p^{2}},\qquad
\text{radius} = r' = \frac{1 + p^{2}}{|1 - p^{2}|}.
$$
The image of the $x$-axis under this transformation is a circle orthogonal to the distorted circle.
If you're looking to plot these circles parametrically or otherwise describe points on the distorted circle corresponding to points on the original, you don't need $\alpha'$; just set $(x, y) = (\cos\alpha, \sin\alpha)$. If it matters, however, you can calculate $\alpha'$ (a little tediously) by expressing $1 - c'$ and $f(z) - c'$ as plane vectors and using the dot product to find the angle they subtend.
