(I) Preliminaries
Firstly let's collect some assumptions, definitions, and facts from the paper you linked to.
Assumption
- $A$, $B$, $C$, and $D$ are real numbers such that $A^2 + B^2 + C^2 \neq 0$.
Definitions
$\Sigma$ is defined to be the unit sphere in the Euclidean space, i.e.
$$
\Sigma := \big\{(x,y,z)\in\mathbb{R}^3 : x^2+y^2+z^2 = 1\big\}.
$$
$N$ is defined to be the northern pole of the unit sphere, i.e. $N := (0,0,1)$.
$S$ is defined to be the stereographic projection, i.e. the function $S:\Sigma\setminus\{N\}\rightarrow\mathbb{R}^2$ satisfying
$$
S\big((x,y,z)\big) = \Big(\frac{x}{1-z},\frac{y}{1-z}\Big)
$$
for every $(x,y,z) \in \Sigma\setminus\{N\}$.
$R$ is defined to be the function $R:\mathbb{R}^2\rightarrow\mathbb{R}^3$ satisfying
$$
R\big((x,y)\big) = \Big(\frac{2x}{x^2+y^2+1}, \frac{2y}{x^2+y^2+1}, \frac{x^2+y^2-1}{x^2+y^2+1}\Big)
$$
for every $(x,y) \in \mathbb{R}^2$.
$\Pi$ is defined to be the plane $\Pi := \big\{(x,y,z) \in \mathbb{R}^3 : Ax + By + Cz + D = 0\big\}$.
$c$ is defined to be the circle obtained from the intersection of the unit sphere with the plane $\Pi$. In other words, $c := \Sigma\cap\Pi$.
We will also use the following definitions, not defined in the paper:
$\operatorname{Id}_\Sigma$ will denote the identity function on $\Sigma$.
$\operatorname{Id}_{\mathbb{R}^2}$ will denote the identity function on $\mathbb{R}^2$.
$L$ is defined to be the set of points $(x,y) \in \mathbb{R}^2$ satisfying the equation resulting from substituting $(1)$ from the paper into $(2)$ from the paper. In other words,
$$
\begin{multline*}
L := \Big\{(x,y)\in\mathbb{R}^2 :\\
A\frac{2x}{x^2+y^2+1} + B\frac{2y}{x^2+y^2+1} +
C\frac{x^2+y^2-1}{x^2+y^2+1} + D = 0\Big\}.
\end{multline*}
$$
Fact
- $S$ and $R$ are inverses, i.e. (a) $R\circ S = \operatorname{Id}_\Sigma$, and (b) $\operatorname{Img} R \subseteq \Sigma\setminus\{N\}$ and $S\circ R = \operatorname{Id}_{\mathbb{R}^2}$. (I will not prove this fact; I leave it for the reader to verify.)
(II) A formal statement of the problem
With these notations, your question can be formalized as follows. Prove that $L = S[c]$.
(III) Proof
We will prove that $L = S[c]$ by showing that $L \subseteq S[c]$ and that $S[c] \subseteq L$.
We first show that $L \subseteq S[c]$. Let $(x,y) \in L$. Define
$$
\begin{align*}
a &:= \frac{2x}{x^2+y^2+1},\\
b &:= \frac{2y}{x^2+y^2+1},\\
c &:= \frac{x^2+y^2-1}{x^2+y^2+1},\\
p &:= (a,b,c).
\end{align*}
$$
It follows from the definition of $R$ that $p = R\big((x,y)\big)$, and therefore it follows from the fact that $\operatorname{Img} R \subseteq \Sigma$, that $p \in \Sigma$. Furthermore, since $(x,y)\in L$, it follows from the definition of $L$ that $Aa+Bb+Cc+D=0$. Therefore, it follows from the definition of $\Pi$ that $p \in \Pi$. Hence $p \in \Sigma\cap\Pi$. Therefore, it follows from the definition of $c$ that $p \in c$. Since, as we saw, $p = R\big((x,y)\big)$, it follows from the fact that $S\circ R = \operatorname{Id}_{\mathbb{R}^2}$ that $(x,y) = S\Big(R\big((x,y)\big)\Big) = S(p)$. Therefore, since $p \in c$ (as we saw), $(x,y) \in S[c]$.
Secondly we show that $S[c] \subseteq L$. Define $\ell := S(q)$ for some arbitrary $q \in c$. Since by the definition of $S$, $\operatorname{Img} S \subseteq \mathbb{R}^2$, we can define $x$ and $y$ to be the first and second coordinates of $\ell$, respectively, so that $x, y \in \mathbb{R}$ and $\ell = (x,y)$. Define $a$, $b$, $c$, and $p$ as above. Then, by the definition of $L$, we need to show that $Aa+Bb+Cc+D=0$, or equivalently, by the definition of $\Pi$, we need to show that $p \in \Pi$. In fact
$$
\begin{align*}
p &= R\big((x,y)\big)\tag{1}\\
&= R(\ell)\tag{2}\\
&= R\big(S(q)\big)\tag{3}\\
&= q\tag{4}\\
&\in c\tag{5}\\
&= \Sigma\cap\Pi\tag{6}\\
&\subseteq\Pi,
\end{align*}
$$
where
- $(1)$ is by the definition of $R$,
- $(2)$ is by the fact, mentioned above, that $\ell = (x,y)$,
- $(3)$ is by the definition of $\ell$,
- $(4)$ is by the fact that $R\circ S = \operatorname{Id}_\Sigma$,
- $(5)$ is by the selection of $q$, and
- $(6)$ is by the definition of $c$.
Q.E.D.
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