Nonlinear Tessellations

Construction

Step 3, part 1 proof

The inverse Q' of Q in C lies on the ray OQ.
The circle U is orthogonal to C and passes through Q, so Q' lies on U.
In fact, Q' is one of the two intersections of the ray OQ and U.
The polar of Q, the locus of all centers of circles passing through Q and orthogonal to C, is the perpendicular bisector of the segment determined by Q and Q'.
This is precisely the line through A and B.
Hence the circle with center A and passing through Q will be orthogonal to C and thus a geodesic for the Poincare disc.

Return to Step 3.