smc

Substituting $y=x^2-a$ into $x^2+y^2=a^2$, we get $$x^2+(x^2-a{)}^2=a^2⟹x^2+x^4-2a{x}^2=0⟹x^2(x^2-(2a-1))=0$$ Since this is a quartic, there are $4$ total roots (counting multiplicity). We see that $x=0$ always has at least one intersection at $(0,-a)$ (and is in fact a double root). The other two intersection points have $x$ coordinates $\pm \sqrt{2a-1}$. We must have $2a-1>0;$ otherwise we are in the case where the parabola lies entirely above the circle (tangent at the point $(0,a)$). This only results in a single intersection point in the real coordinate plane. Thus, we see that $\boxed{a>\frac{1}{2}}$.