If all circles have the same size, this rules out something like the "two circles touching, both inside a third circle" of your example image. In fact, it would rule out the whole "inside" configuration because a circle can never be "inside" another circle of the same size.
I think in order for the 5 official solutions for two circles to be valid, by definition each circle must be of a different radius. Therefore, each additional circle must also be of a different radius:
⭕️ A = 1” radius
⭕️ B = 2” radius
⭕️ C = 3” radius
and so on.
Side note: It would be extremely helpful for mathematics discussions if we could upload multiple images of our “scratch work” into one post, @Vilen .
Duplicates. Am I wrong in thinking that each setup or scenario would have one less number of variations than the previous setup? Due to duplicates.
So for (1) below, you would have a total of 13 variations: 1 for A, B and C separate; 4 variations of B separate from A and C, 4 variations of A separate from B and C, and 4 variations of C separate from A and B. So would (2) below have one less variation than (1), and (3) would have one less variation than (2), and so on?