The Hyperbolic Circle Theorem
If γεD, 0 < r < 1, and tεH(2), then
t(K(γ,r)) = K(t(γ),r).
Proof
By the Origin Lemma, there is an hinversion i mapping γ to 0 (and hence 0 to γ).
By the Lemma and the following remarks, i(C_{r}) = K(γ,r), so that
t(K(γ,r)) 
= t(i(C_{r})) 



= toi(C_{r}) 



= K(toi(0),r) 
by the Lemma again 


= K(t(γ),r) 
as t(0) = γ 


Lemma
For 0 < r < 1, tεH(2)
maps the circle C_{r} = {z : z =r} to the locus
{z : z  t(0)/t(0)*z  1 = r}.
This locus is K(t(0),r).

Corollary
If z, w ε D and tεH(2), then D(t(z),t(w)) = D(z,w).
Proof
Suppose that D(z,w) = r. Then z ε K(w,r).
Applying t, t(z) ε t(K(w,r))
By the Theorem, t(K(w,r)) = K(t(w),r), so t(z) ε K(t(w),r).
Thus D(t(z),t(w)) = r = D(z,w), as required.

