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 hinversion 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.


circumcircles
The CabriJava figure on the right shows an htriangle 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 hcircle is a circle lying within D, so there can be
at most one hcircle 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 hcircle,
so the htriangle PQR does not have a hyperbolic circumcircle.
The cyan hlines are the hbisectors of the hsegments 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 hbisectors meet,
or the three hbisectors are concurrent (at S).
In the former case, the green circle cuts C, so there is no hcircle.
In the latter case, the green circle is within D, so is an hcircle.
The hcentre is S.
An attempt to mimic the euclidean proof of the existence of a
circumcircle is instructive.
The hbisectors appear to have the required property 
There is an hcircle with hcentre T passing through U and V
if and only if T lies on the hbisector of UV.
In euclidean geometry, if P, Q and R are not on a line, then the
perpendicular bisectors of PQ and QR are nonparallel,
so meet in a point.
In hyperbolic geometry, the hbisectors may be ultraparallel.

