Let us take a 2D unit circle $C$ with center in $M=(0,1)$ and fix a point on $C$, say $A=(0,0)$. Now we start to unroll the circle to the right and watch how $A$ moves; firstly, it turns around $M$ on the perimeter of $C$ clockwise; secondly, it moves to the right exactly about the length of the arc that has already been unrolled.
The trajectory of $A$ can be described as $$ Tr(A)=\{(\alpha-\sin\alpha,1-\cos\alpha):\alpha\in[0,2\pi]\} $$ (which looks a bit like the upper half of an ellipse).
Now the question occurs, how $Tr(A)$ can be described explicitly by the graph of a function. Therefore I would try to solve $$ x=\alpha-\sin\alpha $$ for $\alpha$, which, to me, already seems hopeless. Any ideas would be appreciated!
As you already found out from the comment above, the curve you're describing is a cycloid. The Wikipedia page on the subject gives an equation of the form $x=f(y)$ for half of the "hump", but says (without any proof) that an equation of the form $y=f(x)$ is not possible.