What is the stereographic projection of a meridian of…

What is the stereographic projection of a meridian of geographic longitude $\alpha$ — the half great circle running from the south pole to the north pole at fixed longitude $\alpha$?

Step 1 · Parametrize the meridian

A meridian at longitude $\alpha$ consists of points with $\eta=\xi\tan\alpha$ and $\xi\ge0$ when $\alpha\in\left(-\frac\pi2,\frac\pi2\right]$ (the sign condition that picks out one half of the great circle, not both). Equivalently: $\xi=\rho\cos\alpha,\ \eta=\rho\sin\alpha$ with $\rho=\sqrt{1-\zeta^2}\ge0$ as $\zeta$ ranges over $[-1,1)$.

Step 2 · Apply the projection formulas

$$ x=\frac{2\xi}{1-\zeta}=\frac{2\rho\cos\alpha}{1-\zeta}, \qquad y=\frac{2\eta}{1-\zeta}=\frac{2\rho\sin\alpha}{1-\zeta} $$

Since $\rho\ge0$ and $1-\zeta>0$ throughout (excluding the north pole), the quantity $r:=\dfrac{2\rho}{1-\zeta}$ is always $\ge0$. So:

$$ x=r\cos\alpha, \qquad y=r\sin\alpha, \qquad r\ge0 $$

Step 3 · Identify the image as a ray, not a full line

Because $r\ge0$ always (never negative), the image is exactly the ray from the origin in the direction $e^{i\alpha}$:

$$ \boxed{z=re^{i\alpha}, \qquad r\ge0} \qquad\big(\text{equivalently } \arg z=\alpha \text{ together with } z=0\big) $$

At $\rho=0$ ($\zeta=-1$, the south pole) we get $r=0$, i.e. $z=0$ — the ray's endpoint. As $\zeta\to1$ (approaching the north pole), $r\to\infty$.

Step 4 · Why not the full line?

The full line $y=x\tan\alpha$ (both rays $\arg z=\alpha$ and $\arg z=\alpha+\pi$) is the image of the full great circle through both poles at longitude $\alpha$ — which consists of two meridians, at longitudes $\alpha$ and $\alpha+\pi$, glued at the poles. A single meridian (one value of longitude) covers only half of that great circle ($\xi,\eta$ with a fixed sign relative to $\cos\alpha,\sin\alpha$), so it projects to only half of the line — a ray.

Verification

Numerically tracing the meridian $\alpha=0.7$ from the south pole ($\theta=-\frac\pi2$) to near the north pole ($\theta\to\frac\pi2$) gives image points with $\arg z=0.700$ at every sampled latitude and $r$ increasing monotonically from $0$ to large values — confirming the image is exactly the ray $\arg z=\alpha$, never crossing to $\arg z=\alpha+\pi$.