14. (25 points) Let $A$, $B$, and $C$ be distinct points on the parabola $y=x^{2}$, and $R$ be the radius of the circumcircle of $\triangle ABC$. Find the range of values for $R$.

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14. For any $a>0$, take $A\left(a, a^{2}\right), B\left(-a, a^{2}\right), O(0,0)$.
Then the circumcenter of $\triangle O A B$ is on the $y$-axis, denoted as $C(0, b)$.
Since the perpendicular bisector of $O A$ is
$$
y-\frac{a^{2}}{2}=-\frac{1}{a}\left(x-\frac{a}{2}\right) \text {. }
$$

Let $x=0$, we get $b=\frac{1}{2}+\frac{a^{2}}{2}$.
Therefore, the radius of the circumcircle of $\triangle O A B$ can take all values in the interval $\left(\frac{1}{2},+\infty\right)$.

Let $A, B, C$ be different points on the parabola $y=x^{2}$, $R$ be the radius of the circumcircle of $\triangle A B C$, and $(a, b)$ be its center. Then
$$
(x-a)^{2}+\left(x^{2}-b\right)^{2}-R^{2}=0
$$

has three different real roots. Therefore, there must be four real roots, denoted as $x_{1}, x_{2}, x_{3}, x_{4}$.
Rearranging equation (1) gives
$$
\begin{array}{l}
x^{4}+(1-2 b) x^{2}-2 a x+a^{2}+b^{2}-R^{2}=0 \\
\Rightarrow x_{1}+x_{2}+x_{3}+x_{4}=0 \\
\Rightarrow x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+2 \sum_{i \neq j} x_{i} x_{j}=0 . \\
\text { Also } \sum_{i \neq j} x_{i} x_{j}=1-2 b\frac{1}{2} . \\
\text { By } 2(2 b-1)=-2 \sum_{i \neq j} x_{i} x_{j} \\
=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2} \\
\geqslant 4 \sqrt[4]{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}}=4 \sqrt{\left|x_{1} x_{2} x_{3} x_{4}\right|} \\
=4 \sqrt{\left|a^{2}+b^{2}-R^{2}\right|} \\
\Rightarrow 4 b^{2}-4 b+1 \geqslant 4\left(a^{2}+b^{2}-R^{2}\right) \\
\Rightarrow R^{2}-a^{2}+\frac{1}{4} \geqslant b>\frac{1}{2} \\
\Rightarrow R^{2}>a^{2}+\frac{1}{4} \Rightarrow R>\frac{1}{2} .
\end{array}
$$