Let $\gamma:(a,b)\rightarrow\mathbb{R}^2$ be a regular smooth plane curve. Assume that the signed curvature of $\gamma$ at $t_0\in (a,b)$, i.e. $\kappa_*(t_0)$ is nonzero. Then there exists a neighbourhood $U$ of $t_0$ such that for all $\{t_1<t_2<t_3\}\subseteq U$, $\gamma (t_1),\gamma (t_2)$ and $\gamma (t_3)$ does not lie on a straight line.
$\bf{(}$Let us give a proof of this fact: If not, then we can find a sequence $\{t_{1,n}<t_{2,n}<t_{3,n}\}_{n\in\mathbb{N}}$ such that for each $i=1,2,3$, $t_{i,n}\rightarrow t_0$ as $n\rightarrow\infty$, and $\gamma (t_{1,n}),\gamma (t_{2,n})$ and $\gamma (t_{3,n})$ are colinear, i.e. $\frac{\gamma (t_{2,n})-\gamma (t_{1,n})}{t_{2,n}-t_{1,n}}=\frac{\gamma (t_{3,n})-\gamma (t_{2,n})}{t_{3,n}-t_{2,n}}$, i.e. $\dot{\gamma}(\xi_{1,2,n})=\dot{\gamma}(\eta_{2,3,n})$ for some $\xi_{1,2,n}$ in between $t_{1,n}$ and $t_{2,n}$, and $\eta_{2,3,n}$ in between $t_{2,n}$ and $t_{3,n}$. Now $\ddot{\gamma}(t_0)\neq 0$ and $\gamma$ is smooth, implies $\dot{\gamma}$ is locally injective. But then talking $n\in\mathbb{N}$ large enough, the equality $\dot{\gamma}(\xi_{1,2,n})=\dot{\gamma}(\eta_{2,3,n})$ contradicts that local injectivity.$\bf{)}$
Now I want to prove the following:
Problem: There exists an open neighbourhood $U$ of $t_0$ such that the set $\big\{$Radius of circle passing through $\gamma(t_1),\,\gamma(t_2),\,\gamma(t_3)$: $\{t_1<t_2<t_3\}\subseteq U$$\big\}$ is a bounded set.
Note that the signed curvature function $\kappa_*$ for a regular smooth curve $\gamma $ is defined by $\kappa_*=\frac{\ddot{\gamma}\cdot J\dot{\gamma}}{\lVert{\dot{\gamma}}\rVert^3}$, where $J$ is the complex structure on $\mathbb{R}^2$ defined by $J(x,y)=(-y,x)$ for $x,y\in\mathbb{R}$.
I got stuck with this problem while studying the notion of Evolute of a smooth regular curve (In particular Theorem 4.12) from the book 'Modern Differential Geometry of Curves and Surfaces with Mathematica' by Abbena, Salamon and Gray. Thanks in advance for any help.