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1 solution
Area of four small circles: 4 π r 2
Area of large circle with twice the radius: π ( 2 r ) 2 = 4 π r 2
Small circles overlap in four equal areas, red, and miss covering four equal areas, blue.
If the totals are the same, as shown, then the overlaps have to add up to the same area as the area that is missed.
Quarters of equal areas are equal, so the red and blue areas are equal.