Existence of inversion

Mar 2009
419
64
Uptown Manhattan, NY, USA
Let \(\displaystyle p\) and \(\displaystyle q\) be circles with unequal radii. Prove there exists an inversion that maps \(\displaystyle p\) onto \(\displaystyle q\).

So my idea was this:

Proof:

Choose an axis that passes through both centers of the circles (so that the axis contains the diameters of \(\displaystyle p\) and \(\displaystyle q\)). Without loss of generality, let the radius of \(\displaystyle p\) be smaller than the radius of \(\displaystyle q\). Choose a point \(\displaystyle C\) on the constructed axis so that \(\displaystyle p\) is between point \(\displaystyle C\) and \(\displaystyle q\) and that \(\displaystyle CD \cdot CD' = CE \cdot CE'\), where \(\displaystyle DE\) is the diameter of \(\displaystyle p\) and \(\displaystyle E'D'\) is the diameter of \(\displaystyle q\), both of which are the diameters we chose to lie of the constructed axis (where \(\displaystyle E\) lies between \(\displaystyle D\) and \(\displaystyle E'\) and \(\displaystyle E'\) lies between \(\displaystyle E\) and \(\displaystyle D'\)). Then we can verify that \(\displaystyle I_{C, \sqrt{CD\cdot CD'}}\) maps \(\displaystyle p\) onto \(\displaystyle q\). 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.
 

Opalg

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Aug 2007
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Let \(\displaystyle p\) and \(\displaystyle q\) be circles with unequal radii. Prove there exists an inversion that maps \(\displaystyle p\) onto \(\displaystyle q\).

So my idea was this:

Proof:

Choose an axis that passes through both centers of the circles (so that the axis contains the diameters of \(\displaystyle p\) and \(\displaystyle q\)). Without loss of generality, let the radius of \(\displaystyle p\) be smaller than the radius of \(\displaystyle q\). Choose a point \(\displaystyle C\) on the constructed axis so that \(\displaystyle p\) is between point \(\displaystyle C\) and \(\displaystyle q\) and that \(\displaystyle CD \cdot CD' = CE \cdot CE'\), where \(\displaystyle DE\) is the diameter of \(\displaystyle p\) and \(\displaystyle E'D'\) is the diameter of \(\displaystyle q\), both of which are the diameters we chose to lie of the constructed axis (where \(\displaystyle \color{red}E\) lies between \(\displaystyle \color{red}D\) and \(\displaystyle \color{red}E'\) and \(\displaystyle \color{red}E'\) lies between \(\displaystyle \color{red}E\) and \(\displaystyle \color{red}D'\)). Then we can verify that \(\displaystyle I_{C, \sqrt{CD\cdot CD'}}\) maps \(\displaystyle p\) onto \(\displaystyle q\). 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.
That looks broadly correct to me (I haven't checked it carefully), but I am a bit dubious about the statement in red. That seemed to me to imply that the circles p and q intersect each other. If I'm right, then you also need to consider the possibilities that the circles are disjoint, with the smaller one lying either completely inside or completely outside the larger one. In any case, I think that you need to give a bit of explanation of why it is possible to choose C so that the condition \(\displaystyle CD \cdot CD' = CE \cdot CE'\) holds.
 
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Mar 2009
419
64
Uptown Manhattan, NY, USA
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 \(\displaystyle C\), 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).