Characterizing a circle.

Question

Is it correct to characterize a circle by saying that it is a closed curve in $\mathbb{R}^2$ such that all points on the curve are equidistant from a single fixed point?

I am familiar with shapes such as reuleaux polygons, so this isn't what I am getting at (that is shapes of constant width).

My gut feeling tells me that one can construct a (convoluted) curve in $\mathbb{R}^2$ with this property that wouldn't be geometrically (or even topologically) a circle. I particularly would like to stick with closed curves in $\mathbb{R}^2$ and comparing them with $S^1$. I am sure this question might have different answers for higher dimensions, but I am not concerned with these.

I mean formally we would write down that the circle is $S^1 = \{ (x,y) \in \mathbb{R}^2 \, | \, x^2 + y^2 = r \}$, but does this bi-conditionally translate to my first statement?

Edit to this original question: The way I phrased the question was poor in hindsight partly because I hadn't quite pinned down what exactly what I wanted to ask. The more precise statement I am trying to make is the following,

For the metric space $(\mathbb{R}^2 , d)$ where $d$ is the standard Euclidean metric, we define the circle centered at the point $(a,b) \in \mathbb{R}^2$ to be the set $S^1 = \{ (x,y) \in \mathbb{R}^2 \, | \, (x-a)^2 + (y-b)^2 = r^2 \}$, where we say that $r$ is the radius of the circle. Is this definition equivalent to saying a circle is a closed curve in $\mathbb{R}^2$ such that all points on the curve are equidistant from a single fixed point? I do not care for the single point to be the origin and the distant to be $r$

Again, my gut feeling says that one might be able to construct a (convoluted) curve that has this property which is not the same realization as $S^1$.

Answer 1

A circle is the set of all points of distance $r$ from a certain point (its centre). If you have a closed curve such that all points are equidistant from a fixed centre, this (or rather its image) may not be a circle. For example, consider the closed curve $[0,2\pi]\to\mathbb R^2$, $t\mapsto (\sin\sin t,\cos\sin t)$, which merely goes back and forth on some arc of the unit circle.

If you add the condition that the curve be simple, you are fine, though (modulo distinction between curve and image of curve): By the distance condition, your curve is an continuous, injective map (apart from the end points) essentially into a circle and must be onto (and in fact a homeomorphism). This follows from investigating all continuous maps $S^1\to S^1$ and homotopy.

Answer 2

As far as I know, given a topological metric space $(X,d)$ and a point $a\in X$ and $r\in \Bbb R^+$, a circle $(C,a,r)$ is defined as to be $\{b\in X\mid d(a,b)=r\}$.

Since for any real number $r>0$ we have $x^2+y^2=r\iff \sqrt{x^2+y^2}=\sqrt{r}$, it turns out the set you're defining is just a circle in the Euclidean metric about the origin. Of course the set for a circle about the point $(h,k)$ is just $\{(x,y)\mid d((x,y),(h,k))=\sqrt{(x-h)^2+(y-k)^2}=\sqrt{r}\}$.

There are other metrics on $\Bbb R^2$, though. I would invite you to consider what the unit circles for the following metrics would look like:

1. $d((x,y),(w,z))=|x-w|+|y-z|$
2. $d((x,y),(w,z))=\max(|x-w|,|y-z|)$

---

There are probably other characterizations of circles, but I can't think of any less complicated than the one using a distance function at the moment.

---

(Added after new edits)

For the metric space $(\mathbb{R}^2 , d)$ where $d$ is the standard Euclidean metric, we define the circle centered at the point $(a,b) \in \mathbb{R}^2$ to be the set $S^1 = \{ (x,y) \in \mathbb{R}^2 \, | \, (x-a)^2 + (y-b)^2 = r^2 \}$, where we say that $r$ is the radius of the circle. Is this definition equivalent to saying a circle is a closed curve in $\mathbb{R}^2$ such that all points on the curve are equidistant from a single fixed point?

In $\Bbb R^2$ they are virtually identical ("all points at a fixed distance from a fixed point") except that the one you are suggesting has "and the set is a closed curve" tacked on, which is a property of the set in question and doesn't really have to be part of its definition.

In the Euclidean distance, $d(a,b)=r \iff d(a,b)^2=r^2$, so the fact that the square root is missing inside the set you mentioned is inessential.