 Solution

An ideal hyperbolic triangle has its three vertices at infinity, with sides given by geodesics. In curvature $-1$ its angles are zero and its area is $\pi$, using the hyperbolic triangle area formula $\pi -(\alpha +\beta +\gamma )$.

To calculate the area of a hyperbolic disc, use the Poincare disc model centered at the origin. Hyperbolic radial distance is $\rho ={\int }_0^{R}2dr/(1-r^2)=2\mathrm{artanh}R$, so its Euclidean radius is $R=\tanh (\rho /2)$. The metric area density is $4rdrd\theta /(1-r^2{)}^2$, givingEvery hyperbolic triangle, including ones with ideal vertices, has area at most $\pi$. But $A(2)=2\pi (\cosh 2-1)>\pi$. A hyperbolic triangle is geodesically convex, so if it contains a complete hyperbolic circle, it also contains the enclosed hyperbolic disc, its geodesic convex hull. This would violate the area comparison. Therefore no hyperbolic triangle contains a complete circle of radius two.