6. The center of a unit square coincides with the center of a circle, and the square is inside the circle. If a point is randomly selected on the circle, the probability that this point can see two complete sides of the square is $\frac{1}{2}$. Then, the radius of the circle is ( ).
(A) $\frac{\sqrt{6}+\sqrt{2}}{2}$
(B) $\frac{\sqrt{4+2 \sqrt{2}}}{2}$
(C) $\sqrt{6}+\sqrt{2}$
(D) $\frac{\sqrt{4+2 \sqrt{2}}}{4}$
6. B.

On the minor arc between the adjacent sides of the square, you can see the complete two sides. According to the problem, "the probability of seeing two complete sides of the square is $\frac{1}{2}$," which means that the 8 intersection points formed by extending the sides of the square and the circle divide the circumference into 8 equal parts. After calculation, the radius of the circle is $\frac{\sqrt{4+2 \sqrt{2}}}{2}$.