$\min \left[(x_1 - x_2)^2 + \left|12 - \sqrt{1 - x_1^2} - \sqrt{4x_2}\right|^2\right], \forall x_1, x_2 \in \mathbb{R}$ is:
Step-by-Step Solution
Key Concept: Shortest distance between curves occurs along the common normal, and for a circle the normal passes through its center.
The point $(x_1, y_1)$ lies on the circle $x^2 + (y-12)^2 = 1$ and $(x_2, y_2)$ lies on the parabola $y^2 = 4x$. The shortest distance occurs along the common normal. The normal to the parabola $y^2 = 4x$ is $y = mx - 2m - m^3$, which passes through the circle's center $(0, 12)$ giving $m^3 + 2m + 12 = 0$ with solution $m = -2$. The corresponding point on the parabola is $(4, 4)$, and the minimum distance is $\sqrt{4^2 + 8^2} - 1 = 4\sqrt{5} - 1$.
Correct Answer: 2