Radius of Largest Circle Enclosed in Parabolic Region

To find the radius of the largest circle $C$ enclosed within the region $R = \{(x, y) \in \mathbb{R} \times \mathbb{R} : x \ge 0 \text{ and } y^2 \le 4 - x\}$ with its center lying on the $x$-axis, we analyze the constraints on the circle's center $(h, 0)$ and its radius $r$.

The equation of a circle centered at $(h, 0)$ with radius $r$ is: $(x - h)^2 + y^2 = r^2 \implies y^2 = r^2 - (x - h)^2$

For the circle to be entirely contained within the region $R$, two conditions must be satisfied:

1. Non-negativity constraint ($x \ge 0$): The left-most point of the circle must lie in the region $x \ge 0$: $h - r \ge 0 \implies h \ge r$

2. Parabolic region constraint ($y^2 \le 4 - x$): For every point $(x, y)$ on the circle, the condition $y^2 \le 4 - x$ must hold for all $x \in [h - r, h + r]$: $r^2 - (x - h)^2 \le 4 - x$ $r^2 - (x^2 - 2hx + h^2) \le 4 - x$ $x^2 - (2h + 1)x + \left(h^2 + 4 - r^2\right) \ge 0$

Let $g(x) = x^2 - (2h + 1)x + \left(h^2 + 4 - r^2\right)$. Since $g(x)$ is a quadratic function opening upwards, its absolute minimum occurs at its vertex: $x_{\text{min}} = \frac{2h + 1}{2} = h + \frac{1}{2}$

For $r \ge \frac{1}{2}$, $x_{\text{min}} = h + \frac{1}{2}$ lies inside the domain $[h - r, h + r]$. Therefore, to guarantee that $g(x) \ge 0$ for all $x \in [h - r, h + r]$, the minimum value of $g(x)$ must be non-negative: $g\left(h + \frac{1}{2}\right) \ge 0$

Substituting $x = h + \frac{1}{2}$: $\left(h + \frac{1}{2}\right)^2 - (2h + 1)\left(h + \frac{1}{2}\right) + h^2 + 4 - r^2 \ge 0$ $-\left(h + \frac{1}{2}\right)^2 + h^2 + 4 - r^2 \ge 0$ $-\left(h^2 + h + \frac{1}{4}\right) + h^2 + 4 - r^2 \ge 0$ $\frac{15}{4} - h - r^2 \ge 0 \implies h \le \frac{15}{4} - r^2$

Combining this with the first constraint $h \ge r$, the center $h$ must satisfy: $r \le h \le \frac{15}{4} - r^2$

For a valid center $h$ to exist, the lower bound must not exceed the upper bound: $r \le \frac{15}{4} - r^2$ $r^2 + r - \frac{15}{4} \le 0$ $4r^2 + 4r - 15 \le 0$ $(2r + 5)(2r - 3) \le 0$

Since the radius $r > 0$, we have: $2r - 3 \le 0 \implies r \le \frac{3}{2} = 1.5$

Thus, the maximum possible radius of the circle $C$ is $\frac{3}{2}$ (or $1.5$).