= Solution
Use an <isometry> of the <Poincare disc model> to move the centre to zero. A <hyperbolic disc> of radius $\rho$ then has Euclidean radius $R=\tanh(\rho/2)$ by the <hyperbolic distance> formula. The <area element of a surface> of the <Riemannian metric> is $4r\,dr\,d\theta/(1-r^2)^2$, so
$$
\operatorname{Area}=\int_0^{2\pi}\int_0^R\frac{4r}{(1-r^2)^2}\,dr\,d\theta=4\pi\left(\frac1{1-R^2}-1\right).
$$
Therefore \b[the area is independent of the centre], and
$$
\boxed{\operatorname{Area}=4\pi\sinh^2(\rho/2)=2\pi(\cosh\rho-1).}
$$
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