Congratulations James Page of Hethersett High School, Norwich
on your solution to this question. Again there are many possible
methods of showing that the radius of the circle is equal to the
side of the square.
The blue line $OD$ is a diameter of the circle and a line of
symmetry.
$\angle EDC = 150^{\circ}$ and $\triangle ECD$ isosceles so
$\angle CED = \angle CDE =15^{\circ}$.
Now $\angle OED = 30^{\circ}$ so $\angle OEC = \angle OED -
\angle CED = 15^{\circ}$ and hence the red line $EC$ is a line of
symmetry for the quadrilateral $OEDC$ proving that the radius of
the circle, $OC$ is equal in length to the side of the square.