Book 3
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A circle does not cut a circle at more points than two.
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For, if possible, let the circle ABC cut the circle DEF at more points than two, namely B, C, F, H;
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let BH, BG be joined and bisected at the points K, L, and from K, L let KC, LM be drawn at right angles to BH, BG and carried through to the points A, E.
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Then, since in the circle ABC a straight line AC cuts a straight line BH into two equal parts and at right angles, the centre of the circle ABC is on AC. [III. 1, Por.]
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Again, since in the same circle ABC a straight line NO cuts a straight line BG into two equal parts and at right angles, the centre of the circle ABC is on NO.
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But it was also proved to be on AC, and the straight lines AC, NO meet at no point except at P; therefore the point P is the centre of the circle ABC.
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Similarly we can prove that P is also the centre of the circle DEF; therefore the two circles ABC, DEF which cut one another have the same centre P: which is impossible. [III. 5]
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Therefore etc. Q. E. D. The word circle (κύκλος) is here employed in the unusual sense of the circumference (περιφέρεια) of a circle. Cf. note on I. Def. 15.
PROPOSITION 11.
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If two circles touch one another internally, and their centres be taken, the straight line joining their centres, if it be also produced, will fall on the point of contact of the circles.
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For let the two circles ABC, ADE touch one another internally at the point A, and let the centre F of the circle ABC, and the centre G of ADE, be taken; I say that the straight line joined from G to F and produced will fall on A.
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For suppose it does not, but, if possible, let it fall as FGH, and let AF, AG be joined.
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Then, since AG, GF are greater than FA, that is, than FH,
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let FG be subtracted from each; therefore the remainder AG is greater than the remainder GH.
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But AG is equal to GD; therefore GD is also greater than GH, the less than the greater: which is impossible.
