when we have circle in hyperbolic plane,what is the center and radius of this circle in Euclidean plane?

Question

Let C be the hyperbolic circle with center Xi, where x>0 and radius r>0. Find the center and radius of this Euclidean circle.

Answer 1

The distance along the vertical line $x=0$ is obtained by integrating $1/y$. So if your center is $xi$ with $x>0$ and radius is $r>0$, there will be two points on the verticqal line $x=0$ where the hyperbolic circle cuts that line, call them $ai,bi$ where $a<x<b$. Then since hyperbolic distance from $a$ to $x$ is $r$ we have $\int_a^x(1/y)dy=r$, giving $\ln(x)-\ln(a)=r$ which solved for $a$ gives $$a=xe^{-r}.$$ Similarly since hyperbolic distance from $x$ to $b$ is $r$ we obtain $$b=xe^r.$$ Now from treatment of upper half plane model for hyperbolic geometry, a hyperbolic circle becomes a usual circle in the upper half plane, and also geodesics in the upper half plane include vertical lines, so that the two points $ai,bi$ are actually on a diameter of the euclidean circle we want. This means the center of that circle (as a euclidean circle) is obtained on averaging the two points $ai,bi$ to get $$R=(ai+bi)/2=x \cdot \frac{e^r+e^{-r}}{2},$$ which using hyperbolic trig functions is also $x \cosh(r).$ If needed, the euclidean radius comes out $x \sinh(r).$