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The inscribed square has its vertices on the circle. By symmetry, its center coincides with the center of the circle, O. Two neighboring vertices V1 and V2 have distances OV1 = OV2 = 1, and their angle is V1OV2 = 90d. By Pythagoras's theorem, distance V1V2 = sqrt(2).

The circumscribed square has the circle tangent to its edges. Consider two neighboring tangent points, T1 and T2. The tangent line for OT1 is perpendicular to it, and thus parallel to OT2. Likewise, the tangent line for OT2 is parallel to OT1. The intersection of those tangent lines I will call X. Angles T1OT2, XT1O, and OT2X are all 90d, making T2XT1 also 90d. This makes quadrangle T1OT2X a rectangle, and since OT1 and OT2 both have length 1, T1OT2X is a square with all sides having length 1. Since the tangent lines extend in both directions, the distance from one intersection to a neighboring one is thus 1+1 = 2.

Thus, the circumscribed square has side length 2, sqrt(2) times the inscribed square's side length. Taking the square gives the area, meaning that the circumscribed square has an area 2 times that of the inscribed square.

Thus, the ratio of the areas of two consecutive squares in this construction is 2.