Show Geodesics are Diameters on Riemann Metric | Variational Calculus Homework

[MarkovMarakov]

Homework Statement

The Riemann metric on $\{z\in C: |z|<1\}$ is defined as $dx^2+dy^2\over 1-(x^2+y^2)$. I wish to show that the geodesics are diameters. Please help!

Homework Equations

As above. And I suspect the Euler Lagrange equations.

The Attempt at a Solution

I have tried using the Euler Lagrange equations (since one way to do this is clearly to just variational calculus) but I can't get the correct form no matter what :( I even tried changing the metric into polar coordinates but I still don't get the answer.

Also, I have noticed that this metric is very similar to the one for the hyperbolic Poincare disc model, only scaled. Is it possible to directly deduce results from that? Nonetheless I am guessing that the calculus method should be more robust.
Thank you.

 

[lippyka]

A:Let $p$ and $q$ be two points on the circle $\{z\in \mathbb C: |z|=1\}$. We have to show that the shortest path from $p$ to $q$ is a diameter.Let $f : [0,1] \to S^1$ be the parametrized geodesic such that $f(0) = p$ and $f(1) = q$. Then the length of the curve is given by$$L(f):= \int_0^1\sqrt{g_{ij}\dot f^i \dot f^j}dt. $$where $\dot f^i = \frac{d}{dt}f^i$ and $g_{ij}$ is the metric tensor. In this case, the metric tensor is $g_{ij} = \frac{1}{r^2} \delta_{ij}$ where $r=1-|f|^2$ and $\delta_{ij}$ is the Kronecker delta. Then we have$$L(f) = \int_0^1 \frac{1}{1-|f|^2}\sqrt{\dot f_x^2 + \dot f_y^2}dt. $$Let $f(t) = x(t) + iy(t)$, then we have$$L(f) = \int_0^1 \frac{1}{1-x^2-y^2}\sqrt{\dot x^2 + \dot y^2}dt. $$Now, since $f(0)=p$ and $f(1)=q$, let us assume that $x(0) = a$ and $x(1) = b$ and similarly $y(0) = c$ and $y(1) = d$. Now, minimizing $L(f)$ with respect to $x(t)$ and $y(t)$ gives the Euler Lagrange equations$$\ddot x - \frac{2x(\dot x^2+\dot y^2)}{1-x^2-y^2} = 0. $$and$$\ddot y - \frac{2y(\dot x^2