Given a circle of radius R: $x^2+y^2\le R$, find probability of horizontal segment with length $\frac{R}{2}$ lie whole inside this circle. Position of segment's center has uniform distribution in circle.

Okay, I draw circle for $R=1$:

Red circle is circle of radius R, green circle is area where center of segment should be in order to be in circle. So I think, that probability is: $$P=\frac{\pi(\frac{R}{2})^2}{\pi R^2}=\frac{1}{4}$$

But answer in the book is $\frac{(4\pi-3\sqrt3)}{6\pi}$

Where I am wrong?