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. |
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.