Twice as Normal

The parabola equation is $y_1=x^2$ while the circle equation (the upper part) is $y_2 = \sqrt{1 - x^2} + c$ , being $c$ the $y$ coordinate of the circle center.

In the intersection point, the intersection is normal. That means that $\frac{dy_1}{dx} = \frac{-1}{\frac{dy_2}{dx}}$ . Or:

$2x = \frac{-1}{\frac{-x}{\sqrt{1-x^2}}}$

$2x^2 = \sqrt{1-x^2}$

$4x^4 + x - 1 = 0$

Solving for bhaskara, and since the intersection point iss both a solution for the parabola and the circle $x^2 = y = 0.390388203...$ and $x = 0.624810534...$

Pluggin this in the circle equation $y_2 = \sqrt{1 - x^2} + c$ leads to $c = -0.390388203$ . Since the displacement is the absolut value, out answer is $\fbox{0.390388203}$