Can we consider a circle as a function?

Question

I heard that a circle is not a function mathematically speaking, because for some $x$, there are two corresponding $y$. I agree, but if the function takes a $x$ and a $y$, we have only one output per input. So is this: $$ f(x, y) = x^2 + y^2 - r^2 $$ a function, mathematically speaking?

Answer 1

"a circle is not a function" is a clumsy way to say that a circle cannot be represented as the graph of a function $y=f(x)$, as a given abscissa may correspond to two distinct ordinates. Recall that a circle is a geometric entity, while a function is a relation between real numbers, so the two terms are not of the same nature.

From the implicit equation

$$x^2+y^2-r^2=0$$

we draw

$$y=\pm\sqrt{r^2-x^2}$$

and a circle can be represented as the union of the graphs of the two functions $f(x)=\sqrt{r^2-x^2}$ and $g(x)=-\sqrt{r^2-x^2}$.

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Now

$$f(x,y)=x^2+y^2-r^2$$ is indeed a function of two arguments, and the locus of $f(x,y)=0$ is a circle.

You might also be interested in the 3D locus

$$z=f(x,y),$$

which corresponds to a paraboloid of revolution.

Any horizontal cross-section through the paraboloid is a circle.

Answer 2

This is indeed a function, but not of a circle. It forms a paraboloid. The equation $f(x,y)=0$ does describe a circle. A function whose image is a circle would be something like $g: \mathbb{R} \to \mathbb{R}^2:t \mapsto g(t) = (\cos(t),\sin(t))$.

However, a circle is a geometrical object, while a function is not, so you can never really say that a function 'is' a circle.