Proof of the Origin Lemma

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

We draw:
  • the line OP
  • the line PQ perpendicular to OP,
    meeting C at Q.
  • the line QR perpendicular to OQ,
    meeting OP at R.
Let C' be the circle with
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.


  • |RP|/|RQ| = |RQ|/|RO|, so
  • |RP||RO| = |RQ|2.
Thus O and P are inverse to the circle C'. The required h-line is C'nD, shown in red.

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.
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