I asked this question in MSE here: <https://math.stackexchange.com/questions/184524/diametrically-opposite-points-go-to-diametrically-opposite-points-under-stereogr> but I didn’t get an answer. I really need to solve this problem and believe me this is not a homework.


Suppose $P\_1$ and $P\_2$ are two diametrically opposite points of a circle $C$ in the complex plane and suppose $\sigma (z)$ denotes the image of $z\in\mathbb{C}$ on Riemann sphere due to the inverse of the stereographic projection. Then I need to prove that $\sigma (P\_1)$ and $\sigma (P\_2)$ are also diametrically opposite points of the circle $\sigma (C)$. 


Intuitively this is obvious but I need to prove this fact algebraically (i.e. using coordinate geometry ). 



Diametrically opposite points go to diametrically opposite points under stereographic projection
Let $C$ be a round circle in $R^2$ centered at $p$. Let $g$ be a rotation fixing $p$; then, as a Moebius transformation of $R^2\cup \infty$, $g$ has two fixed points $p, \infty$. Let $g^\sigma$ be the congutate of $g$ by $\sigma$. Then $g^\sigma$ is again a Moebius transformation of $S^2$ which preserves the circle $C'=\sigma(C)$ and fixes the points $\sigma(\infty)=N$ and $p'=\sigma(p)$, where $N$ is the north pole of $S^2$. Now, suppose that $g$ has order $2$, i.e., sends each $q\in C$ to the diametrically opposite point. Then $g^\sigma$ also has order $2$. 


Note that saying that $\sigma$ sends opposite points on $C$ to opposite points on $C'$, just means that $g^\sigma$ has the property that $g^\sigma(q)=\hat{q}$ (here $\hat{q}\in C'$ denotes the point diametrically opposite to $q$ on $C'$). Under this assumption, $g^\sigma|C'$ would act as a rigid order 2 rotation $R$ on $C'$. Extend $R$ to the entire $S^2$. Since $R$ agrees with $g^\sigma$ on $C'$, and both are orientation-preserving Moebius transformations, $g^\sigma=R$. In particular, $R$ fixes $N$ and $p'$. Thus, $p'=-N$. However, $\sigma(-N)=0$, thus, $p=0$. Therefore, $C$ has center at the origin. 


This shows that if $C$ is a circle satisfying the property you want, then $C$ is centered at the origin. The converse is clear too: If $p=0$ then $p'=-N$ and, hence, $g^\sigma$ is the rigid rotation of order $2$ around the vertical axis. Clearly, $g^\sigma(C')=C'$. Thus, $g^\sigma$ sends each point on $C'$ to its opposite. 


This proof is intuitive, geometric, rigorous and computation-free. 


