Two vertices of a square lie on a circle of radius r,and two other vertices lie on a tangent to this circle. Then each side of a square is - $$(A)\frac{3r}{2}\quad (B)\frac{4r}{3}\quad (C)\frac{6r}{5}\quad (D)\frac{8r}{5}$$ Source
After drawing the diagram I found that two vertices lie outside the circle and two on the circle ,but I can't tell whether the diagonal of the square passes through the center of the circle. I am stuck here.
If you draw a diameter for the circle from the point of tangency, it bisects the chord connecting the two points of the square that lie on the circle. The chord theorem says
$$\left(s\over2\right)^2=s(2r-s)$$
from which you can find that
$$s={8r\over5}$$