Commentary on Proposition III.12¶
Back to Proposition III.12.
Heron says on III. 11: “Euclid in proposition 11 has supposed the two circles to touch internally, made his proposition deal with this case and proved what was sought in it. But I will show how it is to be proved if the contact is external.” He then gives substantially the proof and figure of III. 12. It seems clear that neither Heron nor an-Nairīzī had III. 12 in this place.
Campanus and the Arabic edition of Naṣīraddīn aṭ-Ṭūsī have nothing more of III. 12 than the following addition to III. 11. “In the case of external contact the two lines ae and eb will be greater than ab, whence ad and cb will be greater than the whole ab, which is false.” (The points a, b, c, d, e correspond respectively to G, F, C, D, A in the above figure.) It is most probable that Theon or some other editor added Heron’s proof in his edition and made Prop. 12 out of it (an-Nairīzī, ed. Curtze, pp. 121—2). An-Nairīzī and Campanus, conformably with what has been said, number Prop. 13 of Heiberg’s text Prop. 12, and so on through the Book.
What was said in the note on the last proposition applies, mutatis mutandis, to this. Camerer proceeds in the same manner as before; and we may use the same alternative argument in this case also.
Euclid’s proof is valid provided only that, if FG, joining the assumed centres, meets the circle with centre F in C and the other circle in D, C is not within the circle ADE and D is not within the circle ABC. (The proof is equally valid whether C, D coincide or the successive points are, as drawn in the figure, in the order F, C, D, G.) Now, if C is within the circle ADE and D within the circle ABC, the circles must have cut between A and C and between A and D. Hence, as before, they must also have another corresponding point common on the other side of CD. That is, the circles must have three common points: which is impossible.
Hence Euclid’s proof is valid if F, A, G form a triangle, and the only hypothesis which has still to be disproved is the hypothesis which he would in any case have left to the reader, namely that A does not lie on FG but on FG produced in either direction. In this case, as before, either C, D must coincide or C is nearer F than D is. Then the radius FC must be equal to FA: which is impossible, since FC cannot be greater than FD, and must therefore be less than FA.
Given the same preliminaries, III. 12 can be proved by means of III. 8.
Again, when the proposition III. 12 is once proved, III. 8 helps us to prove at once that the circles lie entirely outside each other and have no other common point than the point of contact.
Among Pappus’ lemmas to Apollonius’ Tactiones are the two partial converses of this proposition corresponding to those given in the last note. Lemma 4 (VII. p. 824) is to the effect that, if AB, AC be in one straight line, B and C being on opposite sides of A, the circles drawn on AB, AC as diameters touch externally at A. Lemma 3 (VII. p. 822) states that, if two circles touch externally at A and BAC is drawn through A cutting both circles and containing the centre of one, BAC will also contain the centre of the other. The proofs, as before, use the common tangent at A.
Mr H. M. Taylor gets over the difficulties involved by III. 11, 12 in a manner which is most ingenious but not Euclidean. He first proves that, if two circles meet at a point not in the same straight line with their centres, the circles intersect at that point; this is very easily established by means of III. 7, 8 and the third similar theorem. Then he gives as a corollary the statement that, if two circles touch, the point of contact is in the same straight line with their centres. It is not explained how this is inferred from the substantive proposition; it seems, however, to be a logical inference simply. By the proposition, every $A$ (circles meeting at a point not in the same straight line with the centre) is $B$ (circles which intersect); therefore every not-$B$ is not-$A$, i.e. circles which do not intersect do not meet at a point not in the same straight line with the centres. Now non-intersecting circles may either meet (i.e. touch) or not meet. In the former case they must meet on the line of centres: for, if they met at a point not in that line, they would intersect. But such a purely logical inference is foreign to Euclid’s manner. As De Morgan says, “Euclid may have been ignorant of the identity of ‘Every $X$ is $Y$’ and ‘Every not-$Y$ is not-$X$,’ for anything that appears in his writings; he makes the one follow from the other by a new proof each time” (quoted in Keynes’ Formal Logic, p. 81).
There is no difficulty in proving, by means of I. 20, Mr Taylor’s next proposition that, if two circles meet at a point which lies in the same straight line as their centres and is between the centres, the circles touch at that point, and each circle lies without the other. But the similar proof, by means of I. 20, of the corresponding theorem for internal contact seems to be open to the same objection as Euclid’s proof of III. 11 in that it assumes without proof that the circle which has its centre nearest to the point of meeting is the “inner” circle. Lastly, in order to prove that, if two circles have a point of contact, they do not meet at any other point, Mr Taylor uses the questionable corollary. Therefore in any case his alternative procedure does not seem preferable to Euclid’s.
The alternative to Eucl. III. 11—13 which finds most favour in modern continental text-books (e.g. Legendre, Baltzer, Henrici and Treutlein, Veronese, Ingrami, Enriques and Amaldi) connects the number, position and nature of the coincidences between points on two circles with the relation in which the distance between their centres stands to the length of their radii. Enriques and Amaldi, whose treatment of the different cases is typical, give the following propositions (Veronese gives them in the converse form).
If the distance between the centres of two circles is greater than the sum of the radii, the two circles have no point common and are external to one another.
Let O, O' be the centres of the circles (which we will call “the circles O, O'”), r, r' their radii respectively.
Since then OO' $>$ r $+$ r', a fortiori OO' $>$ r, and O' is therefore exterior to the circle O.
Next, the circumference of the circle O intersects OO' in a point A, and since OO' $>$ r $+$ r', AO' $>$ r', and A is external to the circle O'.
But O'A is less than any straight line, as O'B, drawn to the circumference of the circle O [III. 8]; hence all points, as B, on the circumference of the circle O are external to the circle O'.
Lastly, if C be any point internal to the circle O, the sum of OC, O'C is greater than O'O, and a fortiori greater than r $+$ r'.
But OC is less than r: therefore O'C is greater than r', or C is external to O.
Similarly we prove that any point on or within the circumference of the circle O' is external to the circle O.
If the distance between the centres of two unequal circles is less than the difference of the radii, the two circumferences have no common point and the lesser circle is entirely within the greater.
Let O, O' be the centres of the two circles, r, r' their radii respectively (r $<$ r').
Since OO' $<$ r' $-$ r, a fortiori OO' $<$ r', so that O is internal to the circle O'.
If A, A' be the points in which the straight line O'O intersects respectively the circumferences of the circles O, O',
so that O'O $+$ OA, or O'A, is less than O'A', and therefore A is internal to the circle O'.
But, of all the straight lines from O' to the circumference of the circle O, O'A passing through the centre O is the greatest [III. 7]; whence all the points of the circumference of O are internal to the circle O'.
A similar argument to the preceding will show that all points within the circle O are internal to the circle O'.
If the distance between the centres of two circles is equal to the sum of the radii, the two circumferences have one point common and one only, and that point is on the line of centres. Each circle is external to the other.
Let O, O' be the centres, r, r' the radii of the circles, so that OO' is equal to r $+$ r'.
Thus OO' is greater than r, so that O' is external to the circle O, and the circumference of the circle O cuts OO' in a point A.
And, since OO' is equal to r $+$ r', and OA to r, it follows that O'A is equal to r', so that A belongs also to the circumference of the circle O'.
The proof that all other points on, and all points within, the circumference of the circle O are external to the circle O' follows the similar proof of prop. 1 above. And similarly all points (except A) on, and all points within, the circumference of the circle O' are external to the circle O.
The two circles, having one common point only, touch at that point, which lies, as shown, on the line of centres. And, since the circles are external to one another, they touch externally.
If the distance between the centres of two unequal circles is equal to the difference between the radii, the two circumferences have one point and one only in common, and that point lies on the line of centres. The lesser circle is within the other.
The proof is that of prop. 2 above, mutatis mutandis.
The circles here touch internally at the point on the line of centres.
If the distance between the centres of two circles is less than the sum, and greater than the difference, of the radii, the two circumferences have two common points symmetrically situated with respect to the line of centres but not lying on that line.
Let O, O' be the centres of the two circles, r, r' their radii, r' being the greater, so that
It follows that in any case OO' $+$ r $>$ r', so that, if OM be taken on O'O produced equal to r (so that M is on the circumference of the circle O), M is external to the circle O'.
We have to use the same Postulate as in Eucl. I. 1 that
An arc of a circle which has one extremity within and the other without a given circle has one point common with the latter and only one; from which it follows, if we consider two such arcs making a complete circumference, that, if a circumference of a circle passes through one point internal to, and one point external to a given circle, it cuts the latter circle in two points.
We have then to prove that the circle O, besides having one point M of its circumference external to the circle O', has one other point of its circumference (L) internal to the latter circle.
Three cases have to be distinguished according as OO' is greater than, equal to, or less than, the radius r of the lesser circle.
OO' $>$ r. (See the preceding figure.)
Measure OL along OO' equal to r, so that L lies on the circumference of the circle O.
Then, since OO' $<$ r $+$ r', O'L will be less than r', so that L is within the circle O'.
OO' $=$ r.
In this case the circumference of the circle O passes through O', or L coincides with O'.
OO' $<$ r.
If we measure OL along OO' equal to r, the point L will lie on the circumference of the circle O.
Then O'L $=$ r $-$ OO', so that O'L $<$ r, and a fortiori O'L $<$ r', so that L lies within the circle O'.
Thus, in all three cases, since the circumference of O passes through one point (M) external to, and one point (L) internal to, the circle O', the two circumferences intersect in two points A, B [Post.].
And A, B cannot lie on the line of centres OO', since this straight line intersects the circle O in L, M only, and of these points one is inside, the other outside, the circle O'.
Since AB is a common chord of both circles, the straight line bisecting it at right angles passes through both centres, i.e. is identical with OO'.
And again by means of III. 7, 8 we prove that all points except A, B on the arc ALB lie within the circle O', and all points except A, B on the arc AMB outside that circle; and so on.