Observation.—Propositions xi, xii., may both be included in one enunciation as follows:—“If two circles touch each other at any point, the centres and that point are collinear.” And this latter Proposition is a limiting case of the theorem given in Proposition iii., Cor. 4, that “The line joining the centres of two intersecting circles bisects the common chord perpendicularly.”
Suppose the circle whose centre is O and one of the points of intersection A to remain fixed, while the second circle turns round that point in such a manner that the second point of intersection B becomes ultimately consecutive to A; then, since the line OO′ always bisects AB, we see that when B ultimately becomes consecutive to A, the line OO′ passes through A. In consequence of the motion, the common chord will become in the limit a tangent to each circle, as in the second diagram.—Comberousse, Géométrie Plane, page 57.
Cor. 1.—If two circles touch each other, their point of contact is the union of two points of intersection. Hence a contact counts for two intersections.
Cor. 2.—If two circles touch each other at any point, they cannot have any other common point. For, since two circles cannot have more than two points common [x.], and that the point of contact is equivalent to two common points, circles that touch cannot have any other point common. The following is a formal proof of this Proposition:—Let O, O′ be the centres of the two circles, A the point of contact, and let O′ lie between O and A; take any other point B in the circumference of O. Join O′B; then [vii.] O′B is greater than O′A; therefore the point C is outside the circumference of the smaller circle. Hence B cannot be common to both circles. In like manner, they cannot have any other common point but A.
PROP. XIII.—Theorem.
Two circles cannot have double contact, that it, cannot touch each other in two points.
Dem.—1. If possible let two circles touch each other at two points A and B. Now since the two circles touch each other in A, the line joining their centres passes through A [xi.]. In like manner, it passes through B. Hence the centres and the points A, B are in one right line; therefore AB is a diameter of each circle. Hence, if AB be bisected in E, E must be the centre of each circle—that is, the circles are concentric—which [v.] is impossible.
2. If two circles touched each other externally in two distinct points, then [xii.] the line joining the centres should pass through each point, which is impossible.