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.
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 [iii.]; 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 + FO2; and AE2 has been proved equal to CF2. Hence EO2 is equal to FO2; therefore EO is equal to FO. Hence AB, CD are (Def. vi.) equally distant from the centre.