= Solution
Use the curvature-minus-one <Poincare disc model>, whose <Riemannian metric> is
$$
ds^2=\frac{4(dx^2+dy^2)}{(1-x^2-y^2)^2}.
$$
Its <geodesics> are Euclidean diameters and arcs of circles meeting the boundary circle orthogonally. The <hyperbolic area> of a measurable region $D$ is
$$
\operatorname{Area}(D)=\int_D\frac4{(1-x^2-y^2)^2}\,dx\,dy.
$$
An <isometry> can move the center of a circle to the origin. Its hyperbolic radius $r$ and Euclidean radius $\rho$ then satisfy $r=\int_0^\rho2\,du/(1-u^2)=2\operatorname{artanh}\rho$, so $\rho=\tanh(r/2)$. Integrating the <area element of a surface> in <polar coordinates> yields
$$
A(r)=\int_0^{2\pi}\int_0^\rho\frac{4u}{(1-u^2)^2}\,du\,d\theta
=\frac{4\pi\rho^2}{1-\rho^2}
=\boxed{2\pi(\cosh r-1)}.
$$
Integrating the line element around the circle gives
$$
C(r)=\frac{4\pi\rho}{1-\rho^2}=\boxed{2\pi\sinh r},
\qquad A'(r)=2\pi\sinh r=C(r).
$$
A geometric definition of <Pi> valid in this geometry is
$$
\boxed{\pi=\lim_{r\downarrow0}\frac{C(r)}{2r}
=\lim_{r\downarrow0}\frac{A(r)}{r^2}.}
$$
Indeed $\sinh r=r+O(r^3)$ and $\cosh r=1+r^2/2+O(r^4)$. The circumference-to-diameter ratio at a fixed positive radius is not constant, so the limiting qualification is necessary.
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