Stereographic projection gives a map of the celestial sphere, and it is straightforward to use analytical geometry to find the equation of the projection of an arbitrary circle on the sphere. (The projection is also a circle.) This can then be used to tackle various problems in spherical astronomy and spherical trigonometry.
The 'simple method' given first follows that approach. The 'more advanced method' is an application of bilinear transformations of a complex variable. Projected points are described by complex numbers, and rotations of the sphere then correspond to the bilinear transformations. The projection of any circle is found by rotating a similar circle which is parallel to the plane of projection.
These methods are useful for those (like me) who find algebra easier than geometry.

I. Simple method

Take the origin of rectangular Cartesian coordinates at a point on the surface of a sphere of unit radius so that the center of the sphere is at (0, 0, 1). The stereographic projection of the sphere onto the diametral plane z = 0 is obtained by drawing straight lines from the origin to points on the sphere.
Any circle on the sphere is given by the intersection of a plane

(where p is the perpendicular from the center of the sphere onto the plane) with the sphere, whose equation is

The equation of the cone through the origin and this circle is obtained very simply by using the equation of the plane to make the equation of the sphere homogeneous:

The projection is now given immediately by putting z = 1:

Applications. . . The projection of a circle parallel to the plane of projection, of radius sin as in Fig. 2, is given by l=0, m=0, n=1, p = cos :

x^2 + y^2 = tan^2( /2)

This corresponds to the fundamental result for the stereographic projection of a point in the xz-plane:

[Z] . . . . . . . . . . . . x = a tan( /2).

Celestial longitude is measured along the ecliptic from , the vernal equinox (V.E.) in the direction of the red arrow in Fig. 1. Right ascension (R.A.) is measured along the equator from the V.E. in the direction of the black arrow in Fig. 1.

The stereographic projection of the celestial equator is simply

Let be the inclination of the ecliptic, and P the pole of the equator. When the vernal equinox is on the eastern horizon (having hour angle H = 3/2), we apply [P] with l = - sin , m = 0, n = cos , p=0 to give the equation

x^2 + y^2 + 2x tan = 1.

In the same way, the projection of the horizon at a place of latitude l is

The difference is that as the earth rotates the horizon remains fixed but the ecliptic and heavenly bodies rotate about the polar axis in the direction E- ZS-W. When the hour angle of the V.E. is H, the projection of the ecliptic is to be rotated through an angle 3/2 - H in the xy-plane, giving

x^2 + y^2 - 2 tan (x sinH + y cosH) = 1.

Example. We shall now apply these results to an example in Smart:

Show that the obliquity of the ecliptic can be determined by making observation of the sun's declination d at a noon near the summer solstice by means of the formula = d + q^2 sin 2d, where q is one-half the defect from a right angle of the sun's right ascension. [M.T. 1924.]

At noon the sun has y = 0 and by [Z],

(1) x = - tan( /4 - d/2) = tan d - sec d, by elementary trigonometry.

Now, the hour angle of the V.E. is H = /2 (+/-) 2q, where q is a small positive angle (cf. Fig. 3). Hence the equation of the ecliptic gives (for y = 0)

x^2 - 2x tan cos 2q - 1 = 0.

The negative root is x = tan cos 2q - sqrt(tan^2 cos^2 2q + 1). Comparing with (1),
tan d = tan cos 2q, or

tan = tan d sec 2q.

The result follows easily by calculating tan( - d) and approximating for small q.

II. More advanced method

Let (X, Y, Z) be any point on the sphere (Fig. 2) and (x, y) its stereographic projection. Let z be the complex number x + iy. Then by similar triangles in Fig. 2,

Hence if the sphere is rotated through an angle p about the Z-axis, z is transformed into w, where

It is fascinating to see what happens for rotations about the Y-axis. Since X^2 + Y^2 + Z^2 = 1, we have

Hence

       Z + iX    1 - Y      Z + 1 + iX - Y
       ------ =  ------  =  --------------
       1 + Y     Z - iX     Z + 1 - iX + Y

                            1 + iz
                          = --------  =  e(2i arctan z).
                            1 - iz

A rotation of the sphere through an angle q about the Y-axis multiplies the left hand side of this equation by e(iq); hence if z transforms into w, then

and

Applications.
1. Write z* = x - iy, and similarly for any complex number. Then the projection of the equator is
zz* = 1. When the V.E. is at E, the ecliptic is obtained from the equator by a rotation - about the y-axis. If t = tan( /2), then the projection of the ecliptic is

which is the same result as before.

2. The basic formulas of spherical trigonometry may be derived as follows. Given any spherical triangle ABC, choose axes so that B is at z = i. There are two ways of carrying B into C by successive rotations about the axes of Z, Y, Z. Using the transformations found above, the two complex numbers representing the projection of C are equal. From this the desired formulas may be derived. The details are left as an exercise.

References
W. M. Smart, Spherical Astronomy, Cambridge University Press, 4th edn reprinted, 1947, p.57. _____(back to his example)
A. P. Stone, 'On the stereographic projection of the sphere', The Mathematical Gazette 40, 181-184 (1956).

Last updated: 8 March 2008