Inversion Excursion

An Excursion with The Inversion Transformation

Part 1. Solution and remarks related to a problem assigned earlier in the course.

3. D is a circle with center N tangent to a line l at the point O and C is a circle that passes through the N and is tangent to l at O as well.
Suppose P is on l and PN intersect C = {Q}; Q' is on C so that Q'Q is parallel to ON; and {P'} = NQ' intersect l.
Prove: a) P and Q are inverses with respect to the circle D.

a) Solution: Notice that OQ is perpendicular to NQ because <NQO is inscribed in a semicircle. Thus P is the inverse of Q w.r.t. D.

b) P' and Q' are inverses with respect to the circle D.

b) Solution: The same as part a)

c) P and P' are inverses with respect to the circle with center at O and radius ON.

c) Solution: The key here is that NOP ~ P'ON , so that
NO/P'O = PO/NO, and thus PO*P'O = (NO)² and thus P and P' are inverses.

Part 2. Connecting Inversion to Stereographic Projection of the Sphere.
[Hilbert and Cohn-Vossen Section 36]
Consider the sphere resting on a horizontal plane with the highest point labelled N( the "North Pole").

The projection from N to the plane takes P' on the sphere to P on the plane.
[Stereographic projection.]
This mapping that assigns all points except the point N to points on the plane is 1:1 and onto.

If we consider n the plane that is tangent to the sphere at N, then n is parallel to the image plane. Furthermore, of p' is the plane tangent to the sphere at P', then the line NP makes equal angles with n and p and the line of intersection of n and p' lies on the plane that is perpendicular to NP'.

Lemma: Suppose e, p and p' are planes with PP' in e ; p and p' form equal angles with the line PP',; and p and p' intersect in a line that lies in a plane orthogonal to PP'. If r is the line formed by p intersect e and r' is the line formed by p' intersect e, then the angle formed by r and PP' is equal to the angle formed by r' and PP'.

Prop. : If r' is a tangent to the sphere at P' and r is the image of r' on the plane p' then the angle formed by r and PP' is equal to the angle formed by r' and PP'.

Proof: Apply the Lemma to r which is the intersection of the image plane, p, with the plane determined by r' and NP'.

Part 3. Stereographic Projection preserves the measurement of angles, i.e., Stereographic Projection is a conformal transformation.
Proof: The angle formed by curves on the surface of the sphere at point P' is determined by the angle formed by the tangent lines to the curves at P', that is by r' and s' lying in the tangent plane p' at P'. Now considering the corresponding lines r and s in the plane p which meet at P.
By considering the angles these lines make with PP', and the fact that they lie in the tangent plane and the image plane, one can show that the angle between r and s is equal to the angle between r' and s'.

Part 4. Suppose C' is a circle on the surface of the the sphere that does not pass through the point N. Let C be the image of C' on the image plane, p. Then C is a circle.
Proof: Notice that the tangent planes for P' on C' envelope a right circular cone with base C' and vertex D'. Let D be the projection of D' from N onto the image plane p. Now consider the plane NDP. Draw D'P* in the plane NDP that is parallel to DP meeting NP at P*. If P* is different from P', then since P' and P* lie on the cone, D'P' = D'P*. Now using the fact that N'P* is parallel to NP, we have that DP/D'P* = DP/D'P' = DN/D'N. Thus for any P, DP = D'P'*DN/D'N, which is a constant. Thus the points P form a circle with center D and radius DP.

Part 5. Inversion transforms the set of lines and circles in the plane to the set of lines and circles in the plane. This correspondence transforms all circles through the center of the circle of inversion to lines, and all other circles are transformed to circles.

Part 6. If two circles are orthogonal then the inversion of one circle with respect to the other circle will give a circle or line that is also orthogonal to the circle of inversion.