Let and be circles with unequal radii. Prove there exists an inversion that maps onto .
So my idea was this:
Choose an axis that passes through both centers of the circles (so that the axis contains the diameters of and ). Without loss of generality, let the radius of be smaller than the radius of . Choose a point on the constructed axis so that is between point and and that , where is the diameter of and is the diameter of , both of which are the diameters we chose to lie of the constructed axis (where lies between and and lies between and ). Then we can verify that maps onto . Q.E.D.
This seems to make sense to me, but I want to know if this is a viable proof or if I have left some aspects or cases out. If the circles share a common center then the proof for that case is easy so I omitted it. Thanks.
I guess my writing was a bit ambiguous, but I wasn't assuming the circles intersected. I just wanted to emphasize that points closer to the center of inversion get sent "farther" out, so the endpoint of a diameter that is closer to a diameter inside an inversion circle gets sent out farther than the antipodal point would. Also, as far as finding an appropriate , I think that's just solving an equation which I believe will always have a solution (I guess I am assuming too much but I am not sure how to rigorously verify that such a point exists).