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Proposition 5

If two circles cut one another, then they do not have the same center.
Let the circles ABC and CDG cut one another at the points B and C. I say that they do not have the same center.
| For, if possible, let it be E. Join EC, and draw EFG through at random.
Then, since the point E is the center of the circle ABC, EC equals EF. Again, since the point E is the center of the circle CDG, EC equals EG. | I.Def.15
But EC was proved equal to EF also, therefore EF also equals EG, the less equals the greater which is impossible. Therefore the point E is not the center of the circles ABC and CDG.
Therefore if two circles cut one another, then they do not have the same center.
Q.E.D.

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Note that no use was made in the proof of the point B. That means the proof actually shows that if two circles meet, then they do not have the same center, and that covers not only this proposition but the next, too, where the two touch each other.

This proposition is used in III.10 which states that circles cannot intersect at more than two points.

Book III Introduction - Proposition III.4 - Proposition III.6.

© 1996
D.E.Joyce
Clark University