Proof of the Origin Lemma
For any point P, there is an h-inversion mapping P to O and O to P.
If P ≠ Q, then the h-inversion is unique.
If P = O, then we may take as the h-inversion (the restriction of) reflection in the real axis.
Otherwise, we produce a point R and a circle C' with centre R such that
- C' is orthogonal to C, and
- O and P are inverse to C'.
return to the hyperbolic inversion page
Let C' be the circle with
- the line OP
- the line PQ perpendicular to OP,
meeting C at Q.
- the line QR perpendicular to OQ,
meeting OP at R.
centre R, radius |RQ|.
As the radii RQ and OR are perpendicular,
the circles C and C' are orthogonal.
Observe that <OPQ and <OQR
are right angles, and that the
triangles RPQ and RQO have
<ORQ in common.
Thus the triangles are similar.
Thus O and P are inverse to the circle C'.
The required h-line is C'nD, shown in red.
- |RP|/|RQ| = |RQ|/|RO|, so
- |RP||RO| = |RQ|2.
Note that as O and P are inverse, each is mapped to the other by the h-inversion.
Finally, observe that the above argument
works in reverse -
R must lie on OP, and, if Q is the point
where a tangent from R meets C,
then the similarity argument shows that
QP must be perpendicular to OP.
If you move P, the figure shows the construction.
If P is outside C, there is no h-line.