Image of the Riemann-sphere Let $S$ be the Riemann-sphere (the unit sphere in $\mathbb{R}^3$) and $\psi: S \rightarrow \mathbb{C}$ be defined by $$\psi(x_1, x_2, x_3)=\frac{x_1 + ix_2}{1-x_3}.$$ Let $\pi$ be a plane in $\mathbb{R}^3$ such that the intersection with $S$ is not empty. Show that $\psi(\pi \cap S)$ is a line or a circumference in $\mathbb{C}$.
It's easy to see that if the plane has normal vector $(0,0,k)$, the image is a circumference. But i can't generalize the result.
 A: We first note that $\psi$ has an inverse function, given by
$$
\psi^{-1}(x + iy) = \left(\frac{2x}{x^2 + y^2 + 1}, \frac{2y}{x^2 + y^2 + 1}, \frac{x^2 + y^2 - 1}{x^2 + y^2 + 1}\right).
$$
Let now $P$ be a plane in $\mathbb R^3$ given by
$$
P := \left\{ (x_1, x_2, x_3) \in \mathbb R^3 \middle| \sum_{j=1}^3 a_j x_j = b \right\}.
$$
First, we cover the case where the north pole (i. e. the vector $(0, 0, 1)$) is not contained within the plane. This implies that $a_3 \neq b$, since otherwise the north pole would be contained. Let $x \in P \cap S_2$. If $a := (a_1, a_2, a_3)$, we have due to the Cauchy-Schwarz inequality
$$
a \cdot x \le \|a\| \|x\| = \|a\|
$$
and therefore, since $x \in P$,
$$
\sqrt{\sum_{j=1}^3 a_j^2} \ge b \Rightarrow \sum_{j=1}^3 a_j^2 \ge b^2 \\
\Rightarrow a_1^2 + a_2^2 \ge b^2 - a_3^2 \ge (b - a_3)(b + a_3) \\
\Rightarrow \frac{a_1^2 + a_2^2}{(a_3 - b)^2} + \frac{b + a_3}{a_3 - b} \ge 0.
$$
Now if $x + iy$ is contained in $\psi(P)$, then we have
$$
a_1 \frac{2x}{x^2 + y^2 + 1} + a_2 \frac{2y}{x^2 + y^2 + 1} + a_3 \frac{x^2 + y^2 - 1}{x^2 + y^2 + 1} = b.
$$
We want this to be a circle equation, i. e. an equation of the form
$$
(x - \lambda)^2 + (y - \mu)^2 = r
$$
for some $\lambda, \mu \in \mathbb R$ and $r \in \mathbb R_{\ge 0}$. Hence, we group all the terms of $x$ and $y$ together and complete the squares:
$$
a_1 \frac{2x}{x^2 + y^2 + 1} + a_2 \frac{2y}{x^2 + y^2 + 1} + a_3 \frac{x^2 + y^2 - 1}{x^2 + y^2 + 1} = b \\
\Leftrightarrow 2 a_1 x + 2 a_2 y + a_3 x^2 + a_3 y^2 - a_3 = b (x^2 + y^2 + 1) \\
\Leftrightarrow (a_3 - b) x^2 + 2 a_1 x + (a_3 - b) y^2 + 2 a_2 y = b + a_3 \\
\Leftrightarrow x^2 + 2 x \frac{a_1}{a_3 - b} + y^2 + 2 y \frac{a_2}{a_3 - b} = \frac{b + a_3}{a_3 - b} \\
\Leftrightarrow
\left( x - \frac{a_1}{a_3 - b} \right)^2 + \left( y - \frac{a_2}{a_3 - b} \right)^2 = \frac{b + a_3}{a_3 - b} + \frac{a_1^2 + a_2^2}{(a_3 - b)^2}
$$
Now we consider the case $b = a_3$. Then we have
$$
a_1 \frac{2x}{x^2 + y^2 + 1} + a_2 \frac{2y}{x^2 + y^2 + 1} + a_3 \frac{x^2 + y^2 - 1}{x^2 + y^2 + 1} = b \\
\Leftrightarrow 2 a_1 x + 2 a_2 y + a_3 x^2 + a_3 y^2 - a_3 = b (x^2 + y^2 + 1) \\
\Leftrightarrow 2 a_1 x + 2 a_2 y = b + a_3,
$$
which defines a line.
