There are geometrical diagrams in which it is evident* (to a skilled geometer) that two circles' radii have a certain integer ratio.
For example, for the following diagram, it is evident that the ratio of the brown circles' radius to the largest yellow circles' radius is $1:7$. The image, and proof, are found on pages 15 and 23 here, which has many other examples.
I have seen such diagrams with ratios $1:n$ for $n=1,2,3,\dots,10$, but I have never seen one with ratio $1:11$. So my question is:
Is there a geometrical diagram in which it is evident that two circles' radii have ratio $1:11$?
*By "evident", I mean:
- The diagram can only show circles, circular arcs, lines, labels, and geometrical symbols such as right angle markers. No numbers.
- Circles of the same color are congruent.
- If circles look like they are tangent, then they are tangent.
- If a circle has tangent points that look like they are diametrically opposite, then they are so.
- No three circles of the same color can have collinear centres, either as shown, or upon rotation of circles and their contents.
(The last rule serves to avoid trivial answers such as a straight chain of say $11$ circles of the same color that span the diameter of a larger circle. I added the last rule after @Blue's answer, with their approval.)
Elegant answers are preferred. Proofs are optional.
Fun fact
There is a single diagram of circles in which it is evident that the radii have proportions $1,\dfrac12,\dfrac13,\dfrac14,\dfrac15,\dfrac16,\dfrac17,\dfrac18,\dfrac19,\dfrac{1}{10}$.
The equations of the circles are in this desmos graph.
Best Answer
I believe this works:
I use (yellow) half-size circles and (red) third-of-half-size circles —with the "Y" arrangement conveniently allowing me to avoid a collinear chain of three— to determine (blue) five-sixths-size circles. These leave room for one-eleventh-size circles tangent to the boundary.
Taking the boundary to have radius $66$, here's a numerical verification:
Previous Solution
Note: OP has (reasonably) changed the rules to disallow three or more circles of the same size that can be rotated into collinearity, making this answer invalid ... but also making the question more interesting, so I'm okay with that.
It seems to me that you can achieve any $1:n$ ratio by just expressing $n$ in binary and making a stack of appropriate power-of-$2$ circles (rotating sub-circles thereof to prevent any three of the same size from being collinear; with only finitely-many centers to avoid, it's always possible to find a suitable rotation).
Here's $1:11$, noting that $11 = 1 + 2^1 + 2^3$ ...