Area of a hyperbolic disc

In curvature minus one, a hyperbolic disc of radius $R$ has areaIn the Poincare disc this follows by integrating $4r(1-r^2{)}^{-2}drd\theta$ up to Euclidean radius $\tanh (R/2)$.

If a domain of finite hyperbolic area $A$ contains a hyperbolic disc of radius $r$ in curvature $-1$, then $2\pi (\cosh r-1)\le A$, hence $r\le \mathrm{arcosh}(1+A/(2\pi ))$. For an interior point of a hyperbolic triangle, the open disc with radius equal to its distance from the boundary lies inside the triangle. Its area is at most $\pi$ by Gauss-Bonnet theorem, giving the elementary bound $r\le \mathrm{arcosh}(3/2)$; this need not be the sharp triangle inradius bound.

Hyperbolic discs of area $\pi /2$ have radius $\log 2$. Starting at the origin and placing their centers successively along one radial geodesic gives center distance $n\log 4$ and Euclidean coordinate