I am stuck on problem 3.3.2 of Differential Geometry of Curves and Surfaces by Banchoff and Lovett. The problem is:
Let $\vec{x}(s) \colon I \to \mathbb{R}^2$ be the parametrization by arc length of a curve $\mathcal{C}$. Consider a circle that passes through the three points $\vec{x}(s)$, $\vec{x}(s + h_1)$, and $\vec{x}(s + h_2)$. Prove that as $(h_1, h_2) \to (0, 0)$, the limiting position of this circle is precisely the osculating circle to $\mathcal{C}$ at the point $\vec{x}(s)$.
So I am stuck at the beginning: finding some representation for the center $\vec{c} \colon \mathbb{R}^2 \to \mathbb{R}^2$ so that $\vec{c}(h_1, h_2)$ is the center of the circle determined by $\vec{x}(s + h_1)$ and $\vec{x}(s + h_2)$.
Once I have that, though, I think that I might be able to expand using Taylor series (at this point I think we can assume $\vec{x}$ is smooth) so that $$ \vec{x}(s + h_i) = \vec{x}(s) + h_i \vec{T}(s) + \frac{h_i^2}{2}\kappa_g(s) \vec{U}(s) $$ for $i = 1,2$. Then, if we had a formula for $\vec{c}$ in terms of $\vec{x}(s + h_i)$ then we could maybe obtain that as $(h_1, h_2) \to (0,0)$, $$ \vec{c} \to \vec{x}(s) + \frac{1}{\kappa_g(s)} \vec{U}(s), $$ which would be sufficient to get the claim since at this point in the book we know that this is the center of the osculating circle.
In any case, I am stuck on finding a useful formula for $\vec{c}(h_1, h_2)$, other than the obvious fact that $$ \| \vec{c} - \vec{x}(s)\| = \| \vec{c} - \vec{x}(s + h_1)\| = \| \vec{c} - \vec{x}(s + h_2) \|. $$
