Suppose that K is a hyperbolic circle, horocycle or hypercircle,
and that A, B lie on K. Let h denote
inversion in H, the hyperbolic
perpendicular bisector of AB.
(1) H meets K,
(2) h(K) = K,
(3) if K is a horocycle at X, then h(X) = X, and
(4) if K is a hyperbolic circle with centre C, then h(C) = C.
By applying the origin lemma, we can assume that the hyperbolic
mid-point of AB is at O.
Then the hyperbolic segment AB is also the euclidean segment AB,
and the hyperbolic bisector is the diameter H perpendicular to AB.
The circle defining K passes through A and B, so its euclidean centre
lies on the line defining H. It follows that H cuts K. This proves (1).
Now return to the general position. Suppose that H cuts K at P.
Since H is the hyperbolic bisector of AB, and P is on H, h maps
A,B,P to B,A,P, and hence maps the euclidean arc K to itself.
This proves (2).
(3) is straight-forward since X is the unique point where the circle
defining K meets the boundary C, and h fixes K and C.
(4) is simply the observation that, if inversion in H maps a hyperbolic
circle to itself, then the centre must lie on H.