Let $S : \mathbb C_{\infty} \longrightarrow \mathbb C_{\infty}$ be a Möbius transformation. Does $S$ necessarily send $\infty$ to $\infty\ $?
I am actually following Conway's book on Complex Analysis. There I found a result which states as follows $:$
Let $z_1,z_2,z_3$ and $z_4$ be four distinct points in $\mathbb C_{\infty}.$ Then $(z_1,z_2,z_3,z_4)$ is a real number iff all four points lie on a circle.
I understand that if we can somehow prove that every Möbius transformation sends $\mathbb R_{\infty}$ into a circle then we are through. But the author shows that every Möbius transformation $S : C_{\infty} \longrightarrow \mathbb C_{\infty}$ sends every element of $\mathbb R$ to a circle Here I have two doubts $:$
$(1)$ Why do we consider straight lines as circles?
$(2)$ Why is it enough to show that for any Möbius transformation $S$ that $S$ sends $\mathbb R$ to a circle? If we know that $S$ sends every element of $\mathbb R$ to a circle does it always imply that $S$ sends $\infty$ to the same circle?
Any suggestion in this regard will be greatly appreciated. Thanks!