## Comparing two geometric problems

### Proof of the existence of an inversion

Figure 15

We note that there are many inversions that give the desired result. We impose some extra conditions that restrict the set of possible inversions.

If $$A$$ and $$B$$ are the centers of the circles $$a$$ and $$b$$, we choose an inversion with center on the straight line $$AB$$. We can even have it sending $$b$$ to itself: it is enough that the inversion circle $$c_{I}$$ intersects $$b$$ orthogonaly. And, for that, if $$I$$ is the point we want for the center of the inversion, we only have to build a half-circle $$s$$ with diameter $$IB$$: the circle $$c_{I}$$ with center $$I$$, passing through point $$P$$ the intersection point of $$s$$ and $$b$$, intersects $$b$$ orthogonally at $$P$$. When $$I$$ moves along the line, the inverse $$a'$$ of $$a$$, represented by a dashed circle in the figure, moves as well as its center $$C_{a'}$$. Let's see what happens when $$I$$ moves along the line segment $$AA_{1}$$ (open on right). If $$I=A$$, $$C_{a'} (=A)$$ is considerably to the left of $$B$$; and, when $$I$$ goes to $$A_{1}$$, the center $$C_{a'}$$ goes to infinity to the right of $$B$$. Since the function is defined and is continuous in that interval, there must a position for $$I$$ for which the inverse $$a'$$ of $$a$$ has center $$C_{a'}= B$$, i.e., that inversion sends both circles into the two concentric circles $$a'$$ and $$b'(=b)$$ (with center $$B$$). The argument works in a similar way when $$b$$ is inside $$a$$.

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