PROPOSITION 6.

If two circles touch one another, they will not have the same centre.

For let the two circles ABC, CDE touch one another at the point C;

I say that they will not have the same centre.

C F E B D A

For, if possible, let it be F; let FC be joined, and let FEB be drawn through at random.

Then, since the point F is the centre of the circle ABC,

FC is equal to FB.

Again, since the point F is the centre of the circle CDE,

FC is equal to FE.

But FC was proved equal to FB; therefore FE is also equal to FB, the less to the greater: which is impossible.

Therefore F is not the centre of the circles ABC, CDE.
Therefore etc. Q. E. D.

Dependency Graph

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See Commentary on Proposition III.6.