Infimum length of curves

Let the unit disc $\{(x,y): r^2=x^2+y^2<1\}\subset\mathbb R^2$ be equipped with the Riemannian metric $dx^2 +dy^2\over 1-(x^2+y^2)$. Why does it follow that the shortest/infimum length of curves are diameters? I remember doing an optimization course some time ago, but unfortunately all that was learnt has been unlearnt. Or perhaps there is a simpler way?

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This can be done with some variational calculus. So, given the metric

$$ds^2=\frac{dx^2+dy^2}{1-x^2-y^2}$$

the length of a curve will be given by

$$l=\int ds=\int\sqrt{\frac{dx^2+dy^2}{1-x^2-y^2}}.$$

Now, let us parametrize our curve with a parameter $t$ so to have $x=x(t)$ and $y=y(t)$. We will have

$$l=\int ds=\int\sqrt{\frac{\dot x^2+\dot y^2}{1-x^2-y^2}}dt$$

being dot a derivative with respect to time. Now, we can do variational calculus and we take the shortest paths as those having $\delta l=0$. This gives the following Lagrange equations

$$\frac{d}{dt}\left[\frac{1}{\sqrt{1-x^2-y^2}}\frac{\dot x}{\sqrt{\dot x^2+\dot y^2}}\right]=\frac{x\sqrt{\dot x^2+\dot y^2}}{(1-x^2-y^2)^\frac{3}{2}}$$

$$\frac{d}{dt}\left[\frac{1}{\sqrt{1-x^2-y^2}}\frac{\dot y}{\sqrt{\dot x^2+\dot y^2}}\right]=\frac{y\sqrt{\dot x^2+\dot y^2}}{(1-x^2-y^2)^\frac{3}{2}}.$$

The solution of these equations are straight lines $y=x$.

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Thanks, Jon! I am a bit rusty in variational calculus. Can you explain a little bit about the letting $\delta l=0$ bit? I am not sure where the Lagrange equations come from... –  misere Feb 23 '12 at 23:18
Yes, of course. If you have a functional $$S=\int dt L(\dot x, \dot y,x,y)$$ and do variations with respect to $\dot x, \dot y,x,y$, after integration by part and proper boundary conditions, you are left with Lagrange equations that can be written in the generic form $$\frac{d}{dt}\frac{\partial L}{\partial \dot q_i}=\frac{\partial L}{\partial q_i}$$ being ${\bf q}=(x,y)$. –  Jon Feb 23 '12 at 23:27
Thanks again, Jon! –  misere Feb 23 '12 at 23:54