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.

If not,

A

H

Dem. Let O be the centre of APB. Join OP. I say the centre of the smaller circle is in the line OP. 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

D

P

(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.

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.”

GDOO

Suppose the circle whose centre is 0 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 00' always bisects AB, we see that when B ultimately becomes consecutive to A, the line 00' 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.

0 o' A

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 0 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 is, 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 [x1.]. 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,

E

B

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.

Or thus: Draw a line bisecting AB at right angles. Then this line [I., Cor. 1] must pass through the centre of each circle, and therefore [XI. XII.] must pass through each point of contact, which is impossible. Hence two circles cannot have double contact.

This Proposition is an immediate inference from the theorem [XII., Cor. 1], that a point of contact counts for two intersections, for then two contacts would be equivalent to four intersections; but there cannot be more than two intersections [x.]. It also follows from Prop. XII., Cor. 2, that if two circles touch each other in a point A, they cannot have any other point common: hence they cannot touch again in B.

Exercises.

1. If a variable circle touch two fixed circles externally, the difference of the distances of its centre from the centres of the fixed circles is equal to the difference or the sum of their radii, according as the contacts are of the same or of opposite species (Def. Iv.).

2. If a variable circle be touched by one of two fixed circles internally, and touch the other fixed circle either externally or internally, the sum of the distances of its centre from the centres of the fixed circles is equal to the sum or the difference of their radii, according as the contact with the second circle is of the first or second kind.

3. If through the point of contact of two touching circles any secant be drawn cutting the circles again in two points, the radii drawn to these points are parallel.

4. If two diameters of two touching circles be parallel, the lines from the point of contact to the extremities of one diameter pass through the extremities of the other.

PROP. XIV.-THEOREM.

In equal circles-1. equal chords (AB, CD) are equally distant from the centre. 2. chords which are equally distant from the centre are equal.

E

Dem.-1. Let O be the centre. Draw the perpendiculars OE, OF. Join AO, CO. Then because AB is a chord in a circle, and OE is drawn from the centre cutting it at right angles, it bisects it [.]; therefore AE is the half of AB. In like manner, CF is the half of CD; but AB is equal to CD (hyp.). Therefore AE is equal to CF [I., Axiom VII.]. And because E is a right angle, AO2 is equal to AE2 + EO2. In like manner, CO2 is equal to CF2 + FO2; but AO2 is equal to CO2. Therefore AE2 + EO2 is equal to CF2+FO; and AE2 has been proved equal to CF2. Hence EO is equal to FO2; therefore EO is equal to FO. Hence AB, CD are (Def. vI.) equally distant from the centre.

B

2. Let EO be equal to FO, it is required to prove AB equal to CD. The same construction being made, we have, as before, AE2 + EO2 equal to CF2 + FO2; but EO2 is equal to FO2 (hyp.). Hence AE is equal to CF2, and AE is equal to CF; but AB is double of AE, and CD double of CF. Therefore AB is equal to CD.

Exercise.

If a chord of given length slide round a fixed circle-1. the locus of its middle point is a circle; 2. the locus of any point fixed in the chord is a circle.