# How do I prove that if I have a circle of infinite radius, any finite segment of it is equivalent to a straight line?

I haven't done any math in at least 5 years, so please be gentle :D

I have this intuition that if I took a circle of infinite radius, any finite length segment taken from it would be a straight line segment.

I know that a general circle equation in 2d x-y coords is x^2 + y^2 = r^2, but I'm having some troubles coming up with more ideas, because if I use this equation, quite a few things turn infinite, and are freaking me out.

Any clues would be welcome.

P.S. This is not homework, I'm too old for that, and out of school... this is just a random hobby question. Thanks. Please don't down vote! :D

• "a circle of infinite radius" is a curious object. A general equation for a circle of finite radius might be $(x-a)^2+(y-b)^2=r^2$ where the centre is at $x=a,y=b$ while the radius is $r$, but this does not make much sense when any of $a,b,r$ are infinite. Mar 23 '17 at 16:02
• There is no such thing as a circle of infinite radius. Mar 23 '17 at 16:03

If I were approaching this problem, I would pick points A = (0,1) and B = (0,-1), and put the center of the circle at (-$\sqrt{r^2-1}$,0) [with $r$>1]. Then I would determine a formula for an arbitrary point ($x$,$y$) on the arc [with $x$ as a function of $y$ and $r$], and show that $x$ goes to 0 as $r$ increases without bound (with $y$ held fixed).
There is such a thing as a circle of infinite radius, but like the Square character in Flatland you need first to go beyond the usual number system $\mathbb R$ and work instead in the hyperreals ${}^\ast\mathbb R$. Choosing a positive infinite hyperreal $H\in{}^\ast\mathbb R$, one considers a circle ${}^\ast\! S^1$ of radius $H$ centered at the point $(0,H)\in{}^\ast\mathbb {R}^2$. Then the finite part of the circle is infinitely close to the $x$-axis.
More precisely, denoting the finite part $F\subseteq {}^\ast\! S^1$, one obtains that $\text{st}(F)$ is precisely the $x$-axis $\mathbb{R}\times\{0\}\subseteq \mathbb{R}^2$. Here "st" is the standard part function.