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.
