According to the Wikipedia entry about Klein Beltrami disk, I found that the hyperbolic distance between two points P and Q is determined by the following formula :
$$d(P, Q) = \frac{1}{2} \ln \frac{|AQ||PB|}{|AP||QB|}$$
Where A and B are on the euclidian line $(PQ)$ and on the unit Circle. $|AQ| > |AP|$ and $|PB| > |QB|$
Now, I want to know the hyperbolic distance between O the center of unit disk and a point P.
$$d(O, Q) = \frac{1}{2} \ln \frac{|AQ||OB|}{|AO||QB|}$$ Now I think $|OB| = |AO| = 1$ $$d(O, Q) = \frac{1}{2} \ln \frac{|AQ|}{|QB|}$$ Since A, Q and B are aligned and A et B belong to the unit circle :
$|AB| = |AQ| + |QB| = 2$
and $|AQ| = |AO| + |OQ|$
and $|QB| = |OB| - |OQ|$
Considering $|OQ| = r$ I conclude : $$d(O, Q) = \frac{1}{2} \ln \frac{1+r}{1-r}$$ $$d^{-1}(O, Q) = \frac{e^{2r} - 1}{e^{2r}+1}$$
Am I correct ?
All correct (I think I wrote that on wikipedia) $$d(O, Q) = \frac{1}{2} \ln \frac{1+r}{1-r} = \operatorname{artanh} r$$
see also the formulas for tanh and artanh at en.wikipedia.org/wiki/Hyperbolic_function
I have added it to the wikipedia page
ps related:
For the Poincare disk model he hyperbolic distance between O the center of unit disk and a point P an Euclidean distance r away is $d(O, P) = \ln \frac{1+r}{1-r} = 2 \operatorname{artanh} r $.