Hyperbolic Circle <=> Euclidean Circle.

1. Jul 18, 2008

MathematicalPhysicist

I have this question which is rather simple, basically reiterating a general theorem.

Show that S={z in H||z-i|=3/5} is a hyperbolic circle S={w in H| p(w,w0)=r}
for r>0 and find sinh(r/2) and w0.

Now to show that it's hyperbolic is the easy task, I just want to see if I got my calculations correct for sh(r/2) and w0.

Now it's best to move to the unit disk model of poincare by: f(z)=(z-i)/(z+i) because this is an isometry it keeps the same lengths here.
(D={z in C| |z|<1} so f(S)={u in D||f(z)-f(i)|=|(u-0)|=3/5} so it's a unit eulidean circle around zero, now we have the following relationship between this radius and the hyperbolic radius r=log((1+3/5)/(1-3/5))=2log(2) and sh(r/2)=3/4 and to get back w0 we need to use the fact that f(w0)=(w0-i)/(w0+i)=0 thus w0=i.
Is this approach valid?