Key Concepts and Formulas

• Distance between two points: The distance between points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
• Equation of a circle: The equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.
• Area between curves: The area bounded by $y = f(x)$, $y = g(x)$, $x = a$, and $x = b$ is $\int_a^b |f(x) - g(x)| \, dx$.

Step-by-Step Solution

Step 1: Define the fixed and moving circles

The fixed circle $C_1$ has the equation $x^2 + (y-1)^2 = 1$. Therefore, its center is $(0, 1)$ and its radius is $r_1 = 1$. The moving circle $C_2$ has center $(\alpha, \beta)$ and radius $r_2 = \beta$ (since it touches the x-axis and $\beta > 0$).

Step 2: Apply the external tangency condition

Since the two circles touch externally, the distance between their centers is equal to the sum of their radii: $\sqrt{(\alpha - 0)^2 + (\beta - 1)^2} = 1 + \beta$

Step 3: Derive the equation of the locus

Squaring both sides of the equation, we get: $\alpha^2 + (\beta - 1)^2 = (1 + \beta)^2$ $\alpha^2 + \beta^2 - 2\beta + 1 = 1 + 2\beta + \beta^2$ $\alpha^2 = 4\beta$ Replacing $\alpha$ with $x$ and $\beta$ with $y$, we get the equation of the locus L as: $x^2 = 4y$ or $y = \frac{x^2}{4}$.

Step 4: Find the intersection points of the locus and the line

To find the intersection points of the parabola $y = \frac{x^2}{4}$ and the line $y = 4$, we set them equal to each other: $\frac{x^2}{4} = 4$ $x^2 = 16$ $x = \pm 4$ Thus, the intersection points are $(-4, 4)$ and $(4, 4)$.

Step 5: Set up the integral for the area

The area bounded by the parabola $y = \frac{x^2}{4}$ and the line $y = 4$ is given by: $A = \int_{-4}^{4} \left(4 - \frac{x^2}{4}\right) \, dx$

Step 6: Evaluate the integral

$A = \int_{-4}^{4} \left(4 - \frac{x^2}{4}\right) \, dx = \left[4x - \frac{x^3}{12}\right]_{-4}^{4}$ $A = \left(4(4) - \frac{4^3}{12}\right) - \left(4(-4) - \frac{(-4)^3}{12}\right)$ $A = \left(16 - \frac{64}{12}\right) - \left(-16 + \frac{64}{12}\right)$ $A = 16 - \frac{16}{3} + 16 - \frac{16}{3}$ $A = 32 - \frac{32}{3}$ $A = \frac{96 - 32}{3} = \frac{64}{3}$

The area is $\frac{64}{3}$. This contradicts the correct answer provided. Let us look for mistake in the question itself. The question states that the correct answer is $\frac{32\sqrt{2}}{3}$. The locus of the center of the circle is $x^2=4y$.

Let us assume the radius of the fixed circle is $\sqrt{2}$ instead of $1$. In this case, the equation of the fixed circle is $x^2+(y-1)^2=2$. If we solve with this assumption. Then $\sqrt{\alpha^2+(\beta-1)^2} = \sqrt{2}+\beta$. Squaring both sides $\alpha^2 + \beta^2 -2\beta + 1 = 2+2\sqrt{2}\beta + \beta^2$. $\alpha^2 = 2\beta + 2\sqrt{2}\beta + 1 = (2+2\sqrt{2})\beta+1$. In this case, the question will not be on area under the curve. This means that the locus equation we derived is correct.

Let's assume the correct area is $\frac{32\sqrt{2}}{3}$. If we need to get this area, we can rewrite $A=2 \int_0^a (\sqrt{16} - \frac{x^2}{4})dx = \frac{32\sqrt{2}}{3}$

We have derived the locus and the area correctly. The options or the correct answer provided are incorrect.

The correct answer is $\frac{64}{3}$.

Common Mistakes & Tips

• Be careful when squaring equations to eliminate square roots, as it can introduce extraneous solutions. In this case, both sides are non-negative, so squaring is valid.
• Remember to consider the symmetry of the region when setting up the integral. This can simplify the calculation.
• Always double-check your algebraic manipulations and integral calculations to avoid errors.

Summary

The locus of the center of the moving circle is the parabola $x^2 = 4y$. The area bounded by this parabola and the line $y = 4$ is calculated by integrating the difference between the line and the parabola from $x = -4$ to $x = 4$. The calculated area is $\frac{64}{3}$.

Final Answer

The final answer is \boxed{\frac{64}{3}}. The correct option is (C).