Or search by topic
At the points of intersection of two circles, the points satisfy the equations both both circles and so we have a set of two simultaneous equations to solve. Solving these gives at the intersection of $C_1$ and $C_4$ (so for the points $C$ and $D$), $y=\frac{-(\sqrt{5}+1)}{4}$ and at the intersection of $C_1$and $C_5$, $y=\frac{\sqrt{5}-1}{4}$.
But we can now look at the hint, and find we have the same y coordinates for all our points as those of a regular hexagon. But since we are on the circle, we can work out the $x$ coordinates from the $y$ coordinates. So we have the same points as the regular pentagon in the notes section, and so this is a regular pentagon.
Tom from Bristol Grammar School then suggested a method to construct a regular decagon using the pentagon that we've already constructed.