Distance between a point and the origin in Poincaré Disk How to calculate the distance between the origin and the point $p=(a,0)$, $a>0$, using the metric $g = \frac{4}{(1-x^2-y^2)^2}(dx^2+dy^2)$? I don't how to use correctly these $dx^2$'s
 A: You'll have to integrate along the path. $\mathrm dx^2+\mathrm dy^2$ is simply the squared length of an infinitesimally short piece of your path. The geodesic from the origin to a point on the $x$ axis will follow that axis, so you can omit those $y$s and compute the distance as
$$ d
= \int_0^a \sqrt{\frac{4}{\left(1-x^2\right)^2}\mathrm dx^2}
= \int_0^a \frac{2}{1-x^2}\mathrm dx
$$
Actually computing that integral will likely take some work. Wolfram Alpha aparently isn't up to this task. But you don't have to do this work yourself: the Wikipedia article on the Poincaré disk model mentions your formula (for arbitrary dimensions and labeled “metric tensor”), but it also has a formula where you simply plug in the norms of some euclidean vectors:
$$ d(u,v)
= \operatorname{arcosh}\left(1+2\frac{\lVert u-v\rVert^2}{\left(1-\lVert u\rVert^2\right)\left(1-\lVert v\rVert^2\right)}\right) \\
d\left(\begin{pmatrix}0\\0\end{pmatrix},\begin{pmatrix}a\\0\end{pmatrix}\right)
= \operatorname{arcosh}\left(1+2\frac{a^2}{1-a^2}\right) \\
$$
You can compute the derivative of this to check that it agrees with the integrand in the above formula. At least for positive $a$ it does, otherwise you'd need absolute values in that formula.
You can also plug that formula into Wolfram Alpha to find nicer notations. Among those listed for $a>0$ you'll find the following shorter alternative:
$$d = \log\left(\frac{1+a}{1-a}\right) $$
