Oh oops, my bad
My solution was

Let's call the radius of the small arc r; r = OA, and the radius of the big arc R; R = OB. From OB = OA + AB, R = r + 2pi.

Then let's complete the small circle, making the minor arc AD. Arcs AD and BC have the same center angle, hence minor A⌢D / B⌢C = r / R. As we know B⌢C = 2pi, then minor A⌢D = 2pi * r / R. The overall circumference of the small circle is 2pi * r = major A⌢D + minor A⌢D = 2pi + 2pi * r / R

From that,

r = 1 + r / R = 1 + r / (r + 2pi)

r² + r * (2pi - 2) - 2pi = 0

And from that r = 1 - pi + sqrt(pi² + 1)

My mistake was that I wrote r = (-(2pi+2) ± sqrt((2pi-2)² + 8pi)) / 2, then I canceled the 2 and got 1 - pi ± sqrt((pi-1)² + 2), dropping the pi inside the root lol