PROPOSITION 10. THEOREM. [FIRST PROOF.] One circle cannot cut another at more than two points. E H If possible, let DABC, EABC be two circles, cutting one another at more than two points, namely at A, B, C. Construction. Proof. Join AB, BC. Draw FH, bisecting AB at rt. angles; 1. 10, 11. and draw GH bisecting BC at rt. angles. Because AB is a chord of both circles, and because FH bisects AB at rt. angles, ..the centre of both circles lies in FH. III. 1. Cor. Again, because BC is a chord of both circles, and because GH bisects BC at right angles, .. the centre of both circles lies in GH. III. 1. Cor. Hence H, the only point common to FH and GH, is the centre of both circles; which is impossible, for circles which cut one another cannot have a common centre. III. 5. Therefore one circle cannot cut another at more than two points. Q.E.D. COROLLARIES. (i) Two circles cannot have three points in common without coinciding entirely. (ii) Two circles cannot have a common arc without coinciding entirely. (iii) Only one circle can be described through three points, which are not in the same straight line. PROPOSITION 10. THEOREM. [SECOND PROOF.] One circle cannot cut another at more than two points. B E If possible, let DABC, EABC be two circles, cutting one another at more than two points, namely at A, B, C. Construction. Find H, the centre of the O DABC, III. 1. and join HA, HB, HC. Proof. Since H is the centre of the .. HA, HB, HC are all equal. And because H is a point within the DABC, 1. Def. 15. EABC, from which more than two equal st. lines, namely HA, HB, HC are drawn to the Oce, .. H is the centre of the EABC: III. 9. that is to say, the two circles have a common centre H; but this is impossible, since they cut one another. III. 5. Therefore one circle cannot cut another in more than two points. Q.E.D. NOTE. Both the proofs of Proposition 10 given by Euclid are indirect. The second of these is imperfect, because it assumes that the centre of the circle DABC must fall within the circle EABC; whereas it may be conceived to fall either without the circle EABC, or on its circumference. Hence to make the proof complete, two additional cases are required. PROPOSITION 11. THEOREM. If two circles touch one another internally, the straight line which joins their centres, being produced, shall pass through the point of contact. Let ABC and ADE be two circles which touch one another internally at A; let F be the centre of the O ABC, and G the centre of the O ADE. Then shall FG produced pass through A. Construction. For if not, suppose, if possible, FG to pass otherwise, as FGEH. Join FA, GA. FGA, the two sides FG, GA are together but FA = FH, being radii of the ABC: FG, GA are together greater than FH. Take away the common part FG : then GA is greater than GH. But GA=GE, being radii of the C ADE: I. 20. Hyp. Hyp. the part greater than the whole; which is impossible. .. FG, when produced, must pass through A. EXERCISES. Q.E.D. 1. If the distance between the centres of two circles is equal to the difference of their radii, then the circles must meet in one point, but in no other; that is, they must touch one another. 2. If two circles whose centres are A and B touch one another internally, and a straight line is drawn through their point of contact, cutting the circumferences at P and Q; shew that the radii AP and BQ are parallel. PROPOSITION 12. THEOREM. If two circles touch one another externally, the straight line which joins their centres shall pass through the point of contact. B HAK E Let ABC and ADE be two circles which touch one another externally at A; let F be the centre of the and G the centre of the O ADE. Construction, Then shall FG pass through A. ABC, For if not, suppose, if possible, FG to pass otherwise, as FHKG. Join FA, GA. Proof. In the ▲ FAG, the two sides FA, GA are together greater than FG: but FA FH, being radii of the C ABC; and GA .. FH and .. FG must pass through A. I. 20. Hyp. Hyp. EXERCISES. Q.E.D. 1. Find the locus of the centres of all circles which touch a given circle at a given point. 2. Find the locus of the centres of all circles of given radius, which touch a given circle. 3. If the distance between the centres of two circles is equal to the sum of their radii, then the circles meet in one point, but in no other; that is, they touch one another. 4. If two circles whose centres are A and B touch one another externally, and a straight line is drawn through their point of contact cutting the circumferences at P and Q; shew that the radii AP and BQ are parallel. PROPOSITION 13. THEOREM. Two circles cannot touch one another at more than one point, whether internally or externally. If possible, let ABC, EDF be two circles which touch one another at more than one point, namely at B and D. Construction. Join BD; and draw GF, bisecting BD at rt. angles. I. 10, 11. Proof. Now, whether the circles touch one another internally, as in Fig 1 or externally as in Fig 2, because B and D are on the Oces of both circles, .. BD is a chord of both circles: the centres of both circles lie in GF, which bisects BD at rt. angles. III. 1. Cor. Hence GF which joins the centres must pass through a point of contact ; III. 11, and 12. which is impossible, since B and D are outside GF. Therefore two circles cannot touch one another at more than one point. Q.E.D. NOTE. It must be observed that the proof here given applies, by virtue of Propositions 11 and 12, to both the above figures: we have therefore omitted the separate discussion of Fig. 2, which finds a place in most editions based on Simson's text. |