How much of the Earth can you see?

Let $P_C = (x,y) \in \mathcal{C}$ . A corresponding tangent vector can be written as: $(-y,x)$ . Therefore, the line passing through points $P$ and $P_C$ must satify: $(x,y) - (-x_0,0) = \lambda (-y,x)$ for some $\lambda \in \mathbb{R}$ . This gives us two sets of equations. Eliminating $\lambda$ and using that $x^2+y^2=1$ , we find that: $A \equiv \left(-\frac{1}{x_0},\frac{\sqrt{x_0^2-1}}{x_0} \right)$ and $B \equiv \left(-\frac{1}{x_0},- \frac{\sqrt{x_0^2-1}}{x_0} \right)$ What is the arc-length? By symmetry, it is twice the arc-length between, say point $(-1,0)$ and A. We find: $\tan (\theta) = \left| {\frac{y}{x}} \right|$ $\Rightarrow \theta = \tan^{-1} \left( \sqrt{x_0^2-1} \right)$

We then have $\gamma = 2 \theta$ and since the perimeter of the circle is $2 \pi$ , the ratio is: $\frac{\gamma}{2 \pi} = \frac{ \tan^{-1} \left( \sqrt{x_0^2-1} \right)}{\pi}$

Note the expected behaviour at $x_0 \rightarrow 1$ and $x_0 \rightarrow \infty$ .