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.

[Hilbert and Cohn-Vossen Section 36]

Consider the sphere resting on a horizontal plane with the highest point labelled N( the "North Pole").

[Stereographic projection.]

If we consider

Lemma: Suppose

Proof:

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

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

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

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.