TITLE: 20 circles in the plane, all passing through the origin
QUESTION [81 upvotes]: Suppose I draw $20$ circles in the plane, all passing through the origin, but no two tangent at the origin. Also, except for the origin, no three circles pass through a common point. How many regions are created in the plane?

REPLY [21 votes]: What I find amusing, is that you don't need to fill the circumference. In fact you will get the same exact number of sectors in these 2 different ways of placing circles:

But in the first (left) case, reasoning on the number of sectors is much easier. Seems in some way you found an "invariant" for circles.
Now by looking at the left picture you will recognize the pattern for sure, Seems you where looking for a "visualization" rather than a simple algebric proof.

And here it is:
$$3(k-1) + 1$$
Where $k$ is number of circles of course. For $1$ circle it holds. It does not hold for $2$ circles (or for $3$ or $5$). The fact is that I was unlucky in choosing $4$ as starting number of circles.
Well let's see why, just focusing on the external region:

Each added circle will go through 1 extra circle compared to previous circles.
That is nothing more than "half square" made of squares. I don't remember the exact formula for that, but since we are given already the number of circles why not use it directly?
$$\frac{(k-1)(k-2)}2$$
So the total number of sectors inside the circle is:
$$2(k-1) + \frac{(k-1)(k-2)}2 + 1$$
Now this time we got something that holds for any number of circles above $3$.

$0$ Circles: hardcoded $0$ regions
$1$ Circle: hardcoded $1$ region
$2$ Circles: hardcoded $3$ regions
$3$ Circles: $(2\times2) + (1) + 1 = 5$ regions
$4$ Circles: $(2\times3) + (3) + 1 = 10$ regions
$5$ Circles: $(2\times4) + (6) + 1 = 15$ regions

This is not a real proof, but helps visualization which is what OP asked for. :)