PR and QS are perpendicular diameters of a large circle with centre O

Notice that by symmetry, the area of each component of the red shaded regions are equal and similarly for the yellow shaded regions. Let x be the area of a component of the red shaded region and y be the area of a component of the yellow shaded region.

Let r be the radius of one of the smaller circles. Then, the smaller circles have area ?r^2, and the unshaded region within one of these circles is ?r^2 - 2x. Since there are four such unshaded regions, the area of the white region in the figure is 4 * (?r^2 - 2x) = 4?r^2 - 8x.

On the other hand, the area of the white region is the difference between the area of the larger circle and the combined area of the red and yellow shaded regions. The area of the larger region is ?(2r)^2 = 4?r^2 and the combined area of the red and yellow regions is 4x + 4y. Thus, the area of the white region is 4?r^2 - (4x + 4y).

Setting the two expressions for the area of the white region equal, we obtain:

4?r^2 - 8x = 4?r^2 - (4x + 4y)

4?r^2 - 8x = 4?r^2 - 4x - 4y

-8x = -4x - 4y

4y = 4x

Since the area of the red shaded region is 4x and the area of the yellow shaded region is 4y, the ratio of the areas is 4x/4y = 1 or 1 : 1.

Answer: A