# 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."

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:

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 $\mathbb{C}$.
• $R(x,y)$ could be understood as function that maps a point in the plane to a point in the sphere. But at the same time, it is expressing a point of the sphere with only coordinates of a point in the plane. That's what I meant here. As to your second question: it's the reasoning that tells you this. Let's see: You have a circle on the sphere, which is defined by the slicing plane intersecting the latter. You express the plane in terms of coordinates of the plane $x,y$ and finally you see these last coordinates do define a circle. – MASL Nov 20 '15 at 18:47