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G.102 intersecting circles have distinct centers — Heath¶

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

For let the circles ( A B C ), ( C D G ) cut one another at the points B, C;

I say that they will not have the same centre.

For, if possible, let it be E; let / E C / be joined, and let / E F G / be drawn through at random.

Then, since the point E is the centre of the circle ( A B C ),

[G.10]/ E C / is equal to / E F /.

Again, since the point E is the centre of the circle ( C D G ),

/ E C / is equal to / E G /.

But / E C / was proved equal to / E F / also; therefore / E F / is also equal to / E G /, the less to the greater: which is impossible.

Therefore the point E is not the centre of the circles ( A B C ), ( C D G ).

Therefore etc. Q. E. D.

Dependency Graph¶