The answer is yes. Consider the arc of the parabola $y=Ax^2$ inside the circle $x^2+(y-1)^2 = 1$, where we initially assume that $A > 1/2$. This intersects the circle in three points, $(0,0)$ and $(\pm \sqrt{2A-1}/A, (2A-1)/A)$. We claim that for $A$ sufficiently large, the length $L$ of the parabolic arc between $(0,0)$ and $(\sqrt{2A-1}/A, (2A-1)/A)$ is greater than $2$, which implies the desired result by symmetry. We express $L$ using the usual formula for arclength:
$$\begin{align*} L &= \int_0^{\sqrt{2A-1}/A} \sqrt{1+(2Ax)^2} \, dx \\ &= \frac{1}{2A} \int_0^{2\sqrt{2A-1}} \sqrt{1+x^2} \, dx \\ &= 2 + \frac{1}{2A} \left( \int_0^{2\sqrt{2A-1}}(\sqrt{1+x^2}-x)\,dx -2\right), \end{align*}$$
where we have artificially introduced $-x$ into the integrand in the last step. Now, for $x \geq 0$,
$$\sqrt{1+x^2}-x = \frac{1}{\sqrt{1+x^2}+x} > \frac{1}{2\sqrt{1+x^2}} \geq \frac{1}{2(x+1)};$$
since $\int_0^\infty dx/(2(x+1))$ diverges, so does $\int_0^\infty (\sqrt{1+x^2}-x)\,dx$. Hence, for sufficiently large $A$, we have $\int_0^{2\sqrt{2A-1}} (\sqrt{1+x^2}-x)\,dx > 2$, and hence $L > 2$.

Note: a numerical computation shows that one must take $A > 34.7$ to obtain $L > 2$, and that the maximum value of $L$ is about $4.0027$, achieved for $A \approx 94.1$.