jmc

Let $r$ be the radius of the circle. Then we can assume that the graph of one of the parabolas is $y=x^2+r.$

Since $\tan 60^{\circ }=\sqrt{3},$ the parabola $y=x^2+r$ will be tangent to the line $y=x\sqrt{3}.$



This means the equation $x^2+r=x\sqrt{3},$ or $x^2-x\sqrt{3}+r=0$ will have exactly one solution. Hence, the discriminant will be 0, so $3-4r=0,$ or $r=\boxed{\frac{3}{4}}.$