Is every curved line a part of a circle? The title says it all.Does every curved line represent a part of a circle?Is there any formal proof for this?
 A: Every smooth enough ($C^2$ maybe? I forget the exact condition) has, for each point, a radius of curvature. Intuitively the radius of the best fitting circle which is tangent at that point. If the radius of curvature is constant, then good things happen. (Try to prove something in the constant radius of curvature case.) 
In general, the entire curve need not lie on a circle. Just about anything other than a circle will serve as a counter example. $x^2$ as an easy one. 
A: Suppose $P$ is some point on a curve (curved line). It’s definitely possible, and in fact, quite typical, for the part of the curve near $P$ to not be shaped exactly like a circle, even for a short bit. In other words, no matter how you try to draw a circle through $P$ that tries to “match” part of the curve, the circle won’t coincide exactly with the curve close to $P$ except at the one point $P$.
You can get a circle to “hug” the curve at $P$, but if you zoomed in to view the curve and the circle up close, the circle and $P$ would be touching only at a single point, in much the same way that a curve and a straight line can be close (tangent), but only touch at one point.
A: Take $f(x)=x^2$ and suppose that there are $a,b\in \mathbb{R}$ such that
$f(x)=y$ for every $x\in (a,b)$.
Here $y=y_0+\sqrt{r^2-(x-x_0)^2}$ for every $x\in (a,b)$  or   $y=y_0-\sqrt{r^2-(x-x_0)^2}$ for every $x\in (a,b)$.
(both of the sign are not possible otherwise $f(x)$ wouldn't be a function)
So, $f(x)=y\Rightarrow x^2=y_0+\sqrt{r^2-(x-x_0)^2}$ for every $x\in (a,b)$ which s a contradiction.
I had the very same question as yours the last two days.
I am surprised it is not true.
