# Do circles divide the plane into more regions than lines?

In this post it is mentioned that $n$ straight lines can divide the plane into a maximum number of $(n^{2}+n+2)/2$ different regions.

What happens if we use circles instead of lines? That is, what is the maximum number of regions into which n circles can divide the plane?

After some exploration it seems to me that in order to get maximum division the circles must intersect pairwise, with no two of them tangent, none of them being inside another and no three of them concurrent (That is no three intersecting at a point).

The answer seems to me to be affirmative, as the number I obtain is $n^{2}-n+2$ different regions. Is that correct?

Proof: let $$\Lambda\Lambda$$ be a collection of lines, and let $$PP$$ be the extended two plane (the Riemann sphere). Let $$P_1P_1$$ be a connected component of $$P\setminus \LambdaP\setminus \Lambda$$. Let $$CC$$ be a small circle entirely contained in $$P_1P_1$$. Let $$\Phi\Phi$$ be the conformal inversion of $$PP$$ about $$CC$$. Then by elementary properties of conformal inversion, $$\Phi(\Lambda)\Phi(\Lambda)$$ is now a collection of circles in $$PP$$. The number of connected components of $$P\setminus \Phi(\Lambda)P\setminus \Phi(\Lambda)$$ is the same as the number of connected components of $$P\setminus \LambdaP\setminus \Lambda$$ since $$\Phi\Phi$$ is continuous. So this shows that for any collection of lines, one can find a collection of circles that divides the plane into at least the same number of regions.
Remark: via the conformal inversion, all the circles in $$\Phi(\Lambda)\Phi(\Lambda)$$ thus constructed pass through the center of the circle $$CC$$. One can imagine that by perturbing one of the circles somewhat to reduce concurrency, one can increase the number of regions.