Again, since Y is a circle and P a point, from which three equal lines PA, PB, PC can be drawn to its circumference, P must be the centre of Y . Hence X and Y are concentric, which [v.] is impossible.
Cor.—Two circles not coinciding cannot have more than two points common. Compare I., Axiom x., that two right lines not coinciding cannot have more than one point common.
PROP. XI.—Theorem.
If one circle (CPD) touch another circle (APB) internally at any point P, the line joining the centres must pass through that point.
Dem.—Let O be the centre of APB. Join OP. I say the centre of the smaller circle is in the line OP. If not, let it be in any other position such as E. Join OE, EP, and produce OE through E to meet the circles in the points C, A. Now since E is a point in the diameter of the larger circle between the centre and A, EA is less than EP [vii. 2]; but EP is equal to EC (hyp.), being radii of the smaller circle. Hence EA is less than EC; which is impossible; consequently the centre of the smaller circle must be in the line OP. Let it be H; then we see that the line joining the centres passes through the point P.
Or thus: Since EP is a line drawn from a point within the circle APB to the circumference, but not forming part of the diameter through E, the circle whose centre is E and radius EP cuts [vii., Cor. 2] APB in P, but it touches it (hyp.) also in P, which is impossible. Hence the centre of the smaller circle CPD must be in the line OP.
PROP. XII.—Theorem.
If two circles (PCF, PDE) have external contact at any point P, the line joining their centres must pass through that point.
Dem.—Let A be the centre of one of the circles. Join AP, and produce it to meet the second circle again in E. I say the centre of the second circle is in the line PE. If not, let it be elsewhere, as at B. Join AB, intersecting the circles in C and D, and join BP. Now since A is the centre of the circle PCF, AP is equal to AC; and since B is the centre of the circle PDE, BP is equal to BD. Hence the sum of the lines AP, BP is equal to the sum of the lines AC, DB; but AB is greater than the sum of AC and DB; therefore AB is greater than the sum of AP, PB—that is, one side of a triangle greater than the sum of the other two–which [I. xx.] is impossible. Hence the centre of the second circle must be in the line PE. Let it be G, and we see that the line through the centres passes through the point P.
Or thus: Since BP is a line drawn from a point without the circle PCF to its circumference, and when produced does not pass through the centre, the circle whose centre is B and radius BP must cut the circle PCF in P [viii., Cor. 3]; but it touches it (hyp.) also in P, which is impossible. Hence the centre of the second circle must be in the line PE.