a circle graph is not a function? I'm a little confused by the rule: If you draw a vertical line that intersects the graph at more than 1 point then it is not a function.
Because then a circle like $y^2 + x^2 = 1$ is not a function? 
And indeed if I rewrite it as $f(x) = \sqrt(1 - x^2)$ then wolfram alpha doesn't draw a circle. I guess I'm missing the intuition as to why this is though?

If you want to have a function that "draws" a circle with radius $r$ and center $P = (x_0, y_0)$ on the cartesian plane, you can use the function $f : [0, 2\pi] \rightarrow \mathbb{R} \times \mathbb{R}$ defined by $$f(\varphi) = (x_0 + r \cos \varphi, y_0 + r \sin \varphi)$$
But, of course, this is not a function from $\mathbb{R}$ to $\mathbb{R}$.
Also, you can define a curve in the plane by means of an equation of two variables $x$ and $y$. If you have a (continuous) function $f : A\subseteq \mathbb{R}\rightarrow \mathbb{R}$, you can get an equation $y = f(x)$ from it, which defines a curve.
But you cannot always transform an equation containing two variables to an equivalent equation $y = f(x)$. The equation $x^2 + y^2 = r^2, r\in\mathbb{R}$ is an example of this fact.
