Properties of Stereographic Projection (using coordinates)

The two most important properties of Stereographic Projection with center N from a sphere S to a plane F are these:

1.      The image of a circle is a circle (or a line).  If c is a circle on S, then the image of c is a circle if N is not on c or is a line if N is on c.

2.      Angle measure is preserved.  If c and d are circles on S, then the angle between the images c' and d' is the same as the angle between c and d on the sphere.

Proof that the image of a circle is a circle (or a line):

This proof will use coordinates and algebra.  There are other more geometric proofs.  We will see one later when we study inversive geometry.

Step 1.  Formula for stereographic projection inverse from plane to the sphere

We can choose coordinates (x,y,z) so that the plane has equation z = 0; the sphere E is the sphere with radius 1 and center (0,0,0); and N = (0,0,1). Given Q = (u, v, 0) in the plane, the inverse stereographic projection is the point P that is the intersection (different from N) of line NQ with the sphere.

Step 2.  Equation for a circle on the sphere

Since every circle on the sphere is the intersection of the sphere with a plane, the points of the circle are the points of the sphere that satisfy the equation of the plane.

The general equation of a plane in 3-space is Ax + By + Cz = D.

The stereographic projection of the circle is the set of points Q for which P = s^-1(Q) is on the circle, so we substitute the formula for P into the equation for the circle on the sphere to get an equation for the set of points in the projection.

P = (1/(1+u² + v²)[2u, 2v, u² + v² - 1] = [x, y, z].

To make the writing simpler, let us temporarily denote u² + v² by M.

Then P = (1/1+M)[2u, 2v, M-1]

Step 3. Substituting into the equation for the plane, we get

(2Au/(1+M)) + (2Bv/(1+M)) + (C(M-1)/(1+M)) = D, then

2Au + 2Bv - C +CM = D + DM

(C-D)M + 2Au + 2Bv -(C+D) = 0, which is

(C-D)(u² + v²) + 2Au + 2Bv -(C+D) = 0.

Step 4.  Conclusion

If C = D, this is the equation of a line.  If C is not equal to D, this is the equation of a circle with center [A/(D-C), B/(D-C)].

Now back to the geometry for a final interpretation.  If C = D, then [0,0,1] is a solution of the equation of the plane, so this means that N is on the plane and thus on the circle.  This coincides what is geometrically clear: If the plane passes through N, then the projection of the points on the plane all lie on the intersection of the plane and the base plane z=0. This says that the projection of a circle c through N is a line.

If C is not equal to D, then  [0,0,1] is a not solution of the equation of the plane, so this means that N is not on the plane nor is N on the circle.  This says that for a circle c not through N, the image of c is a circle.

Outline of Proof that Stereographic projection preserves angles

Suppose that two circles c and d intersect at P on the sphere.  The key to this proof is that we have the same angle of intersection between c' and d' if we replace c and d by two other circles c' tangent to c at P and d' tangent to d at P.  In particular we can choose c' and c' so that they both pass through N as well as P. (This requires some thought to check that this is true, but we will assume it for the moment.

For this c' and d' the images in the plane are two lines m and n through Q, the stereographic image of P.  The line m is parallel to the tangent line to c' at N; and the line n is parallel to the tangent line to d' at N. Thus the angle of intersection of the lines at Q is equal to the angle of intersection of the circles at N. And this angle is equal to the angle of intersection of the circles at P.

There are a number of details to check to make this proof complete, but this is the outline of a proof.