Answer
\[
\frac{1}{2 r}=a
\]
Work Step by Step
Given a circle with radius $r$ and center $(0, r)$ and $a x^{2}=y$ for $x \geq 0$
Along the circle $\frac{1}{r}=\kappa$ and along $a x^{2}=y$ we have $y^{\prime}=2 a x$ and $y^{\prime \prime}=2 a$. Thus:
\[
\begin{aligned}
&\frac{\left|y^{\prime \prime}\right|}{\left(y^{\prime 2}+1\right)^{3 / 2}}=\kappa \\
&=\frac{2 a}{\left[4 a^{2} x^{2}+1\right]^{3 / 2}}
\end{aligned}
\]
And then $2 a=\kappa(0),$ so $\kappa$ is continuous at $0=x$ if
\[
\frac{1}{r}=2 a \quad \rightarrow \quad \frac{1}{2 r}=a
\]
