I think you can replace the whole analysis with a single diagram. You don't actually need to find the closed-form solution in terms of elliptic integrals; the only part we use is whether the integrals have a certain monotonicity property (smaller radius => smaller mean distance). And you can rearrange the "mean distance" integrals in way that the rearranged integrands are pointwise monotonic; and the proof that they're monotonic is an elementary geometric one.
The integral over the circle can be rewritten as an integral of a sum of two terms over a semicircle – the term for the local point, plus the term for the mirrored point on the other semicircle. This sum-term is an everywhere-monotonic function of the radius of the circle (in the proof diagram: XA + XB > XC + XD).
(XAQB, XCRD, and XC¹QD¹ are constructed as parallelograms. XC¹RD¹ doesn't mean anything; it's just a construction whose perimeter compares easily against the other two).
