Circle Inversion

We begin with a circle C of radius r in the Euclidean plane. We will "reflect" over C.

How does CI(C) map a point A? Consider the line through O and A, where O is the origin of C. CI(C) maps A to another point, A', on this line with the following property: . That is, the distance from the origin to A times the distance from the origin to A' is equal to the radius squared. And O, A and A' are CO linear. CI is defined in the entire Euclidean plane; thus A' maps to A. Likewise, CI(C) maps B to B' and B' to B.

Observe that CI conserves angles. That is, AOB = A'OB'. This example illustrates angle conservation at the origin; keep in mind that angles are preserved everywhere in the disk, not just the origin. Such a map is called conformal.

In addition, as the circle inversion isometry is given by Moebius transformations we know that the map takes lines and circles to lines and circles.

As we can see, CI exchanges the interior of the circle with the exterior of the circle, while the points on the circle are not moved at all. Furthermore, observe that CI maps the center of the circle of inversion to infinity, and the "point at infinity" is mapped to the center of the circle. For the sake of clarity we shall refer to the circle we are inverting across as C from now on.

There is an on-line animation of circle inversion. In the animation we are inverting/reflecting the red circle through the black circle. The pink circle is the image of the red circle under the inversion. Notice that the pink circle becomes a line as the red circle hits the origin of the black circle. (The point that touches the origin is mapped to infinity.)

Clearly, circle inversion is defined in the entire Euclidean plane, if we include the point at infinity (otherwise the center is mapped to an undefined point). However, as we are studying CI as an hyperbolic isometry we are interested in only the Poincaré disk. Specifically, we are looking for those CIs that map the disk to the disk.

How do we identify these circle inversions? Inversions over those circles C that intersect the Poincaré disk P orthogonally will map P onto itself. This statement is derived from the fact that CI preserves angles. Consider the following figure.

The circles P and C intersect orthogonally. Thus CI(C) will map P to itself.We have labeled the arc of C inside P, c. Note: If the line we are reflecting over is a diameter, the "circle" inversion is a Euclidean reflection.

Recall that "lines" in the Poincaré model of hyperbolic space are Euclidean arcs of circles (i.e. c) that intersect the boundary of the disk orthogonally. Thus the analogy between circle inversion in hyperbolic space and reflection in Euclidean space is very appropriate. Both functions take a point on the "line" to itself and map all points on one "side" of the "line" to the other "side".

The remaining three isometries of hyperbolic space;

• hyperbolic,
• elliptic,
• parabolic,

Equations

Up: Isometries

Created: Jul 15 1996 --- Last modified: Jul 15 1996