How can I show that circles in the complex plane correspond to circles on the Riemann sphere? How about lines?

Suppose $ T \subset \mathbb{C} $. Show that the corresponding set $ S \subset \Sigma $ is

a. a circle if $ T $ is a circle.
b. a circle minus (0, 0, 1) if $ T $ is a line.

Here we are defining $ \Sigma $ to be the Riemann sphere, given by the set: $$ \Sigma = \left \{(\xi, \eta, \zeta) : \xi^{2} + \eta^{2} + (\zeta - \frac{1}{2})^{2} = \frac{1}{4} \right \} $$

To take a point from $ \mathbb{C} $ to $ \Sigma $ we can use the following:

$$ \xi = \frac{x}{x^{2} + y^{2} + 1}; \eta = \frac{y}{x^{2} + y^{2} + 1}; \zeta = \frac{x^{2} + y^{2}}{x^{2} + y^{2} + 1} $$

We define a circle on $ \Sigma $ to be the intersection of a plane of the form $ A\xi + B\eta + C\zeta = D $ with $ \Sigma $. We also know the converse of this problem is true, that the intersection above yeilds a set in $ \mathbb{C} $ with the following property:

$ (C - D)(x^{2} + y^{2}) + Ax + By = D $. As you can see, when C = D, then an equation for a line is yeilded, otherwise it is a circle.

I really am at a loss about how to solve this problem. The only thing I can think to do is to pick 3 points on a circle or radius $ r $ with center $ z_{0} $, use these points to find two vectors in $ \Sigma $, take their cross product to get a normal vector, use this normal vector to get a plane. Once I have the plane in form $ A\xi + B\eta + C\zeta = D $ then I could prove that the circle I had chosen corresponds exactly with $ (C - D)(x^{2} + y^{2}) + Ax + By = D $. Is there not an easier, less computation way to do this?

I know this is an old question, but here is a proof with essentially zero calculations.

[I'll distinguish the Plane and Riemann Sphere from other planes and spheres by capitalization.]

Let $p$ be the projection mapping from the Plane to the Riemann Sphere. Then $p$ is also an inversion about the north pole that maps the origin of the plane to the south pole. (This can be easily proven by similar triangles.) Note that any inversion maps any sphere not passing through the centre to a sphere. (To see why first prove that the inversion at $O$ that preserves $P$ such that $OP$ is tangent to a sphere also preserves the sphere, because any other point $Q$ on the sphere is mapped to the other intersection of $OQ$ with the sphere, since the plane through $OPQ$ intersects the sphere in a circle.) Now take any circle $C$ in the Plane. $p$ maps any sphere that contains $C$ and does not pass through the north pole to a sphere, and hence $p$ maps $C$ to the intersection of some sphere with the Sphere, which must be a circle!

Also note that $p$ maps any line (a generalized circle) in the Plane to a circle on the Sphere because the plane through the north pole and the line intersects the Sphere in a circle.

Exactly the same method shows the converse. Take any circle $C$ on the Sphere. If $C$ passes through the north pole, it lies on a plane through the north pole, which $p^{-1}$ preserves, and hence $p^{-1}$ maps $C$ to the intersection of that same plane with the Plane, which is a line. If $C$ does not pass through the north pole, it lies on some sphere that does not pass through the north pole, which $p^{-1}$ maps to a sphere, and hence $p^{-1}$ maps $C$ to the intersection of some sphere with the Plane, which is a circle.

The type of map you describe is called a stereographic projection:

http://en.wikipedia.org/wiki/Stereographic_projection

There are many resources out there which proves the thing you are looking for:

See http://www.geom.uiuc.edu/docs/doyle/mpls/handouts/node33.html for a proof that it maps circles to circles for a slightly different sphere, this will give you the basic idea