Stereographic projection (Theorem that circles on the sphere get mapped to circles on the plane) I'm trying to understand the proof of the theorem (given in the link) that states
"Stereographic projection maps circles of the unit sphere, which do not contain the north pole, to circles in the complex plane."
Link to the proof
In the proof it states "In order to obtain an
equation for the projection points (x, y) ∈ C of the circle c under stereographic projection, we substitute (1) into Equation (2), which yields"
Why does plugging in the pre image of points from the image plane into an arbitrary plane give me an equation of the points under stereographic projection? What is the significance of using the pre image?
This is an idea for your use:

 A: Remember, all transformations have two interpretations, a passive and and active one. 
He's looking to express the equation for the plane, which is given in coordinates of $\Sigma$ (that is, $x,\,y,\,h$), in terms of coordinates of the complex $2d$ plane (that is, $x,y$). 
Right before he states that $R(x,y)$ is precisely the function that does such a thing, namely, expressing a point from $\Sigma$ in terms of $x,y$.  
In doing so, the equation of the plane, which involves $x,y,h$ becomes an equation for just the two coordinates of the plane $x,y$. This allows him to check what kind of geometric object we have within $\mathbb{C}$.
A: (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.

Related posts

*

*In this answer I proved that non-degenerate circles in $\mathbb{R}^2$ are mapped by the inverse of the stereographic projection to non-degenerate circles on the unit sphere that do not pass through the northern pole.


*In this answer I proved that lines in $\mathbb{R}^2$ are mapped by the inverse of the stereographic projection to non-degenerate circles on the unit sphere that pass through the northern pole.
