= 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 \b[$\pi$], 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 $\rho=\int_0^R2\,dr/(1-r^2)=2\operatorname{artanh}R$, so its Euclidean radius is $R=\tanh(\rho/2)$. The metric area density is $4r\,dr\,d\theta/(1-r^2)^2$, giving
$$
A(\rho)=2\pi\int_0^R\frac{4r}{(1-r^2)^2}\,dr=4\pi\left(\frac1{1-R^2}-1\right)=\boxed{2\pi(\cosh\rho-1)}.
$$
Every <hyperbolic triangle>, including ones with ideal vertices, has area at most $\pi$. But $A(2)=2\pi(\cosh2-1)>\pi$. 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 \b[no hyperbolic triangle contains a complete circle of radius two].
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