Let $\alpha$ be a unit-speed curve.Then there exists a unique circle $\beta$ such that $\beta(0)=\alpha(0), \ \beta'(0)=\alpha'(0), \ \beta''(0)=\alpha''(0).$
Attempt: Consider $\beta(s)= \textbf{p} +R\ \text{cos}(\frac{s}{R})\textbf{v}_1+R\ \text{sin}(\frac{s}{R})\textbf{v}_2,$ where $\textbf{v}_1, \textbf{v}_2$ are orthonormal vectors. This means I need to show $\beta(s)$ lies in the plane spanned by $\textbf{T}, \ \textbf{N}.$ How do I relate $\beta$ to $\alpha?$
Thank you.
Clearly $R(s):=1/|\kappa(s)|$ is the radius of the circle, let $n(s):=\mathrm{sgn}(\kappa(s))\boldsymbol{N}$, thus $p(s):=\alpha(s)+R(s)n(s)$ is it's center. So there's only one chance to define the circle, namely $$\beta(t)=p(s)+R(s)\Bigl(\sin\bigl(t/R(s)\bigr)\boldsymbol{T}(s)-\cos\bigl(t/R(s)\bigr)n(s)\Bigr).$$
Just calculate $\beta'$ and $\beta''$, write out the given relationships, and solve for $\mathbf p$, $R$, $\mathbf v_1$, $\mathbf v_2$.
Also, note that $\mathbf v_1$ and $\mathbf v_2$ are related -- one is formed by rotating the other by 90 degrees, so they are not independent.
Note that this approach is a brute force mechanical one. I'm using it only because it's the approach you chose. The methods suggested in the other answers are smarter.
Thinking about it geometrically, knowledge of $\alpha'(0)$ and $\alpha''(0)$ allows us to determine the center of curvature of the plane curve at the point $\alpha(0)$. The center of curvature is precisely the center of the osculating circle, and the distance between it and the point $\alpha(0)$ is the radius. Therefore the circle is uniquely determined.
