In Texas kids learn about the number “pi” in 6th grade. I do not know how it is in Kansas. But we are not in Kansas anyway, so we can learn it now.

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We are not in Texas, and we are not in Kansas. We are on the unit disk with hyperbolic geometry defined by the transitive action of SU(1,1) – as it was discussed in the recent series of posts. We know the line element:

$$ds^2=\frac{4(dx^2+dy^2)}{(1-x^2-y^2)^2}.$$

We can now ask this pressing question: what is the ratio of the circumference of a circle to its radius?

Any point of the disk is as good as any other, but in the standard coordinates $(x,y)$ that we are using the origin of the disk is most convenient. Let us introduce polar coordinates

\begin{eqnarray}
x&=&\rho\cos\phi,\\
y&=&\rho\sin\phi.
\end{eqnarray}

We need to express our line element in terms of $(\rho,\phi).$ We calculate
$dx =d\rho\cos \phi-\rho\sin\phi d\phi,$
$dy =d\rho \sin\phi+\rho\cos\phi d\phi,$
$dx^2= d\rho^2\cos^2\phi-2\rho\sin\phi\cos\phi d\rho d\phi+\rho^2\sin^2\phi d\phi^2,$
$dy^2= d\rho^2\sin^2\phi+2\rho\sin\phi\cos\phi d\rho d\phi+\rho^2\cos^2\phi d\phi^2,$
$dx^2+dy^2=d\rho^2+\rho^2 d\phi^2,$
$ds^2=4\frac{d\rho^2+\rho^2d\phi^2}{(1-\rho^2)^2}.$
Consider now a circle with $\rho=\rho_0$ and $\phi$ running from $0$ to $2\pi.$ What is the length $C$ of its circumference? We have to integrate $ds.$ Going around circumference $\rho$ is constant, so $d\rho=0.$ Therefore
$C=\int_0^{2\pi}ds=\int_0^{2\pi}\frac{2\rho_0 d\phi}{1-\rho_0^2}=\frac{4\pi \rho_0}{1-\rho_0^2}.$
What is the length $r$ of its radius? Along the radius $\phi$ is constant, so $d\phi=0$, so
$r=\int_0^{\rho_0} ds=\int_0^{\rho_0}\frac{2d\rho}{1-\rho^2}=2\mathrm{arctanh}\rho_0.$
From the last equation we get
$\rho_0=\tanh\frac{r}{2}.$
Substituting into the formula for $C$:
$C=4\pi\frac{\sinh(r/2)/\cosh(r/2)}{1-\sinh^2(r/2)/\cosh^2(r/2)}=2\pi\frac{2\sinh(r/2)\cosh(r/2)}{\cosh^2(r/2)-\sinh^2(r/2)}.$

Thus
$$C=2\pi \sinh r,$$
and

$$\frac{\mathrm{circumference}}{\mathrm{radius}}=\frac{C}{r}=2\pi\frac{\sinh r}{r}.$$

We can find this formula in Wikipedia article Hyperbolic geometry, in the section Circles and disks (putting there the curvature $R=-1$).

Looking at the graph of the function $\sinh r/r$

we see that the ratio circumference to radius on the disk is always greater than $2\pi.$ In fact the ratio is growing faster and faster with increasing $r$.