Tortilla wrap

The circle area is 20π/3.

Let’s solve for the general case where the circle has radius r, the fold has length 2x and the distance centre to tangency point is d. Now the brown circular segment is part of a circle of equal radius r. The two centres and the intersection points form a rhombus with side r as shown.

Now, applying the Pythagorean theorem in triangle AEB gives 4y²=d²+r². From the four congruent right triangles in the rhombus we have x²+y²=r². Eliminating y from these two equations gives r²=(d²+4x²)/3. The answer comes from filling in d=2 and x=2.