Step-by-Step Solution

1. Understand the Region R

The region R is defined by two conditions: (i) $x \ge 0$: The region is on the right side of the y-axis (in the first and fourth quadrants). (ii) $y^2 \le 4 - x$: This can be written as $x \le 4 - y^2$. The boundary is the parabola $y^2 = 4 - x$, which is equivalent to $y^2 = -(x - 4)$. This is a parabola that opens to the left with its vertex at (4, 0). So, R is the region bounded by the y-axis (x=0) and the parabola $y^2 = 4-x$.

2. Define the Family of Circles F

The circles in the family F must satisfy two properties: (i) They are contained in the region R. (ii) Their centers are on the x-axis.

Let the equation of a circle in F be $(x - h)^2 + y^2 = r^2$, where (h, 0) is the center and r is the radius. Since the center must be in R, we have $h \ge 0$.

3. Formulate Conditions for the Circle to be in R

For a circle $(x - h)^2 + y^2 = r^2$ to be contained in R, it must satisfy:

• Condition A: $x \ge 0$ for all points on the circle. The leftmost point of the circle is at x = h - r. So, we must have $h - r \ge 0$, which implies $h \ge r$.

• Condition B: The circle must be inside the parabola $y^2 = 4 - x$. The circle with the largest radius, C, will be tangent to the boundary parabola $y^2 = 4 - x$. Let's find the condition for tangency.

4. Find the Tangency Condition

To find where the circle and parabola meet, we substitute $y^2 = 4 - x$ into the circle's equation: $(x - h)^2 + (4 - x) = r^2$ $x^2 - 2hx + h^2 + 4 - x = r^2$ $x^2 - (2h + 1)x + (h^2 - r^2 + 4) = 0$

For the circle to be tangent to the parabola, this quadratic equation in x must have exactly one real solution (a double root). This means the discriminant (D) must be zero. $D = (-(2h + 1))^2 - 4(1)(h^2 - r^2 + 4) = 0$ $(4h^2 + 4h + 1) - (4h^2 - 4r^2 + 16) = 0$ $4h + 1 + 4r^2 - 16 = 0$ $4r^2 = 15 - 4h$ $r^2 = \frac{15 - 4h}{4}$ This equation relates the radius r and the center's x-coordinate h for any circle in F that is tangent to the parabola.

5. Maximize the Radius

Our goal is to find the circle C with the largest radius r. We need to maximize r, or equivalently, $r^2$. From the relation $r^2 = (15 - 4h)/4$, we can see that to maximize $r^2$, we need to minimize h.

Now, we use Condition A: $h \ge r$. Since both h and r are non-negative, we can square the inequality: $h^2 \ge r^2$.

Substitute the expression for $r^2$: $h^2 \ge \frac{15 - 4h}{4}$ $4h^2 \ge 15 - 4h$ $4h^2 + 4h - 15 \ge 0$

To find when this inequality holds, we first find the roots of the corresponding quadratic equation $4h^2 + 4h - 15 = 0$. Using the quadratic formula, $h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $h = \frac{-4 \pm \sqrt{4^2 - 4(4)(-15)}}{2(4)}$ $h = \frac{-4 \pm \sqrt{16 + 240}}{8} = \frac{-4 \pm \sqrt{256}}{8} = \frac{-4 \pm 16}{8}$ The roots are $h_1 = \frac{12}{8} = \frac{3}{2}$ and $h_2 = \frac{-20}{8} = -\frac{5}{2}$.

Since the parabola $4h^2 + 4h - 15$ opens upwards, the inequality $4h^2 + 4h - 15 \ge 0$ is satisfied for $h \le -5/2$ or $h \ge 3/2$. We also know that the center (h, 0) must be in R, so $h \ge 0$. Combining these conditions, we must have $h \ge 3/2$.

To maximize $r^2$, we must choose the minimum possible value for h, which is h = 3/2.

6. Find the Point of Tangency ($\alpha$, $\beta$)

For the circle C with the largest radius, its center is at h = 3/2. The point $(\alpha, \beta)$ is where this circle C meets the curve $y^2 = 4 - x$. The x-coordinate, $\alpha$, is the repeated root of the quadratic equation we found in Step 4: $x^2 - (2h + 1)x + (h^2 - r^2 + 4) = 0$

First, let's find $r^2$ for this circle: $r^2 = \frac{15 - 4h}{4} = \frac{15 - 4(3/2)}{4} = \frac{15 - 6}{4} = \frac{9}{4}$

Now, substitute h = 3/2 and $r^2 = 9/4$ into the quadratic equation: $x^2 - (2(\frac{3}{2}) + 1)x + ((\frac{3}{2})^2 - \frac{9}{4} + 4) = 0$ $x^2 - (3 + 1)x + (\frac{9}{4} - \frac{9}{4} + 4) = 0$ $x^2 - 4x + 4 = 0$ $(x - 2)^2 = 0$

The only solution is x = 2. This is the x-coordinate of the point of tangency. Therefore, $\alpha = 2$.

For completeness, we can find $\beta$: $\beta^2 = 4 - \alpha = 4 - 2 = 2$ $\beta = \pm \sqrt{2}$ The points of tangency are $(2, \sqrt{2})$ and $(2, -\sqrt{2})$.

7. Final Answer

The value of $\alpha$ is 2.