1

There are 1 best solutions below

3
On

The radius of the first circle is easily found with Pythagore (half of the square's diagonal). This radius is also the half of the length of the second square, thus. You find the radius of the second circle with Pythagore again, and so on: $$c_1 = 1$$ $$r_1 = \frac{1}{2}\sqrt{c_1^2 + c_1^2} = \frac{\sqrt{2}}{2}c_1$$ $$c_2 = 2.r_1 = 2.\frac{\sqrt{2}}{2}c_1=\sqrt{2}c_1$$ $$r_2 = \frac{1}{2}\sqrt{c_2^2 + c_2^2} = \frac{\sqrt{2}}{2}c_2 = \frac{2}{2}c_1 = c_1$$ $$c_3 = \ ...$$