Experiments with circles

Theorem 3
If K is a circle lying within D, and tεH(2),
then t(K) is a circle lying within D.

Although circles map to circles, centres do not in general map
to centres, and the radius may change.

The CabriJava figure on the right allows you to experiment.
It shows an circle K with centre O, passing through the point Q.
The green circle K' is the image under the h-inversion taking O to P.
The point R is the centre of K'.
You can move Q to change the size of K,
and move P to alter the transformation.

Note that O maps to P, not to the centre R of K'.
As P approaches the boundary C, the green circle
shrinks, but remains inside C.
You may also observe that O, P and R are collinear,
and that, as Q approaches O, K' approaces P.

K and hence K' vanish when Q = O.


The CabriJava figure on the right shows an h-triangle PQR.
Standard euclidean geometry contains the result that, three
points are either collinear or lie on a unique circle.
The green circle is the euclidean circle through P, Q and R.
An h-circle is a circle lying within D, so there can be
at most one h-circle through P, Q and R.

By moving P, Q or R, you should be able to create figures where
the green circle cuts C. In such a case, it is not an h-circle,
so the h-triangle PQR does not have a hyperbolic circumcircle.

The cyan h-lines are the h-bisectors of the h-segments PQ, QR and RP.
These are the hyperbolic analogues of the perpendicular bisectors of the
sides of the euclidean triangle PRQ.
By experimenting, you should observe that
either no two h-bisectors meet,
or      the three h-bisectors are concurrent (at S).
In the former case, the green circle cuts C, so there is no h-circle.
In the latter case, the green circle is within D, so is an h-circle.
The h-centre is S.

An attempt to mimic the euclidean proof of the existence of a
circumcircle is instructive.
The h-bisectors appear to have the required property -
There is an h-circle with h-centre T passing through U and V
if and only if T lies on the h-bisector of UV.
In euclidean geometry, if P, Q and R are not on a line, then the
perpendicular bisectors of PQ and QR are non-parallel,
so meet in a point.
In hyperbolic geometry, the h-bisectors may be ultraparallel.

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