# How to find a circle which approximates an unit speed curve at the origin?

A parametrization of a circle by arc length may be written as $$\gamma(t)=c+r\cos(t/r)e_1+r\sin(t/r)e_2.$$ Suppose $\beta$ is an unit speed curve such that its curvature $\kappa$ satisfies $\kappa(0)>0$.

How to show there is one, and only one, circle which approximates $\beta$ in near $t=0$ in the sense

$$\gamma(0)=\beta(0), \gamma^{'}(0)=\beta^{'}(0)\quad \textrm{and}\quad \gamma^{''}(0)=\beta^{''}(0).$$

I suppose we must use Taylor's formula, but I wasn't able to solve this problem.

• Curvature is the reciprocal of this circle's radius. Of course the tangent is easily found. – hardmath Apr 19 '14 at 13:06

## 1 Answer

The parametrization of circle you stated assumes that $\gamma'(t)$ is parallel to $e_2$ at time $t=0$. So, if $\beta'(0)$ is not parallel to $e_2$, you have a problem. This nuisance can be avoided by writing $$\gamma(t)=c+r\cos((t+a)/r)e_1+r\sin((t+a)/r)e_2 \tag{1}$$ with constant $a\in\mathbb R$. The Taylor formula says $$\gamma(t) = \big(c+r\cos (a/r) e_1 + r \sin (a/r) e_2 \big) + \big( -\sin(a/r) e_1+\cos(a/r) e_2 \big) t + \big(-\cos(a/r) e_1-\sin(a/r) e_2\big)r^{-1}t^2 +O(t^3)$$ Comparing the above to $$\beta(t) = \beta(0) + \beta'(0) t + \frac12 \beta''(0)t^2 +O(t^3)$$ we observe that

1. $r^{-1}=\frac12|\beta''(0)|$, so $r$ is determined. (Unless $\beta''(0)=0$, when the radius of curvature is infinite and there is no circle you are looking for.)
2. Since $\beta'(0)$ is a unit vector, there exists a unique $a$ (up to an integer multiple of $2\pi r$) such that $\beta'(0) =-\sin(a/r) e_1+\cos(a/r) e_2$.
3. $c$ is uniquely determined from $\big(c+r\cos (a/r) e_1 + r \sin (a/r) e_2 \big) = \beta(0)$.