Solution:

We can square both sides of the second curve to get $x^{2} = \frac{y^{3}}{1 - y}$, which further rearranges to
$$\frac{x^{2}}{(x^{2} + y^{2})^{2}} = \frac{y}{x^{2} + y^{2}}.$$
This relation implies that curves $y = x^{2}$ and $x^{2} = \frac{y^{3}}{1 - y}$ map to each other under inversion about the unit circle $x^{2} + y^{2} = 1$. Therefore, the unique circle we seek must be invariant under inversion about $x^{2} + y^{2} = 1$.

Since the circle is tangent to $y = x^{2}$, we know that the circle is of the form
$$x^{2} + (y - y_{0})^{2} = r^{2}.$$
We know that the length of the tangent from $(0,0)$ to this circle is $1$. Since the distance from $(0,0)$ to the center of the circle is $y_{0}$, using the Pythagorean Theorem gives $r^{2} + 1 = y_{0}^{2}$.

Because the parabola $y = x^{2}$ is tangent to this circle at two distinct points, the equation $x^{2} + (x^{2} - y_{0})^{2} = r^{2} = y_{0}^{2} - 1$ must have two double roots. Therefore,
$$x^{2} + (x^{2} - y_{0})^{2} - (y_{0}^{2} - 1) = x^{4} - (2y_{0} - 1)x^{2} + 1$$
must be a perfect square, so $2y_{0} - 1 = 2 \implies y_{0} = \frac{3}{2}$.

This means $r^{2} = y_{0}^{2} - 1 = \frac{5}{4}$, so the radius of the circle is $\left[\frac{\sqrt{5}}{2}\right]$.