Key Concepts and Formulas

• Differential Equations: Separable differential equations can be solved by separating variables and integrating.
• Equation of a Circle: The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$. The area of a circle is $A = \pi r^2$.
• Equation of a Normal to a Circle: The normal to a circle at a point on its circumference passes through the center of the circle. The slope of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$. The point-slope form of a line is $y - y_1 = m(x - x_1)$.

Step-by-Step Solution

Step 1: Solve the differential equation. We are given the differential equation $\frac{dy}{dx} + \frac{x+a}{y-2} = 0$. We need to solve this equation to find the equation of the curve $C$.

$\frac{dy}{dx} = -\frac{x+a}{y-2}$ $(y-2)dy = -(x+a)dx$ Integrating both sides: $\int (y-2)dy = \int -(x+a)dx$ $\frac{y^2}{2} - 2y = -\frac{x^2}{2} - ax + C'$ Multiplying by 2: $y^2 - 4y = -x^2 - 2ax + 2C'$ $x^2 + 2ax + y^2 - 4y = 2C'$ Completing the squares for $x$ and $y$: $(x^2 + 2ax + a^2) + (y^2 - 4y + 4) = 2C' + a^2 + 4$ $(x+a)^2 + (y-2)^2 = a^2 + 4 + 2C'$ Let $R^2 = a^2 + 4 + 2C'$. Then the equation becomes: $(x+a)^2 + (y-2)^2 = R^2$ This is the equation of a circle with center $(-a, 2)$ and radius $R$.

Step 2: Use the initial condition y(1) = 0. We are given that $y(1) = 0$. Substituting $x=1$ and $y=0$ into the equation of the circle: $(1+a)^2 + (0-2)^2 = R^2$ $(1+a)^2 + 4 = R^2 \quad \ldots (1)$

Step 3: Use the area information to find R and a. We are given that the area enclosed by the curve $C$ is $4\pi$. Since the curve is a circle, its area is given by $A = \pi R^2$. Therefore, $\pi R^2 = 4\pi$ $R^2 = 4$ Substituting $R^2 = 4$ into equation (1): $(1+a)^2 + 4 = 4$ $(1+a)^2 = 0$ $1+a = 0$ $a = -1$ So, the equation of the circle is $(x-1)^2 + (y-2)^2 = 4$. The center is $(1, 2)$ and the radius is $2$.

Step 4: Find the points P and Q where the curve intersects the y-axis. To find the points where the circle intersects the y-axis, we set $x=0$: $(0-1)^2 + (y-2)^2 = 4$ $1 + (y-2)^2 = 4$ $(y-2)^2 = 3$ $y-2 = \pm \sqrt{3}$ $y = 2 \pm \sqrt{3}$ The points of intersection are $P(0, 2+\sqrt{3})$ and $Q(0, 2-\sqrt{3})$.

Step 5: Find the equations of the normals at P and Q. Since the normals to a circle pass through the center, the normal at $P$ passes through $(0, 2+\sqrt{3})$ and $(1, 2)$. The slope of the normal at $P$ is: $m_P = \frac{2 - (2+\sqrt{3})}{1-0} = -\sqrt{3}$ The equation of the normal at $P$ is: $y - (2+\sqrt{3}) = -\sqrt{3}(x-0)$ $y = -\sqrt{3}x + 2 + \sqrt{3}$

The normal at $Q$ passes through $(0, 2-\sqrt{3})$ and $(1, 2)$. The slope of the normal at $Q$ is: $m_Q = \frac{2 - (2-\sqrt{3})}{1-0} = \sqrt{3}$ The equation of the normal at $Q$ is: $y - (2-\sqrt{3}) = \sqrt{3}(x-0)$ $y = \sqrt{3}x + 2 - \sqrt{3}$

Step 6: Find the x-intercepts R and S of the normals. To find the x-intercepts, we set $y=0$:

For the normal at $P$: $0 = -\sqrt{3}x + 2 + \sqrt{3}$ $\sqrt{3}x = 2 + \sqrt{3}$ $x = \frac{2 + \sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3} + 3}{3} = \frac{2\sqrt{3}}{3} + 1$ So, $R = (\frac{2\sqrt{3}}{3} + 1, 0)$.

For the normal at $Q$: $0 = \sqrt{3}x + 2 - \sqrt{3}$ $-\sqrt{3}x = 2 - \sqrt{3}$ $x = \frac{-2 + \sqrt{3}}{\sqrt{3}} = \frac{-2\sqrt{3} + 3}{3} = 1 - \frac{2\sqrt{3}}{3}$ So, $S = (1 - \frac{2\sqrt{3}}{3}, 0)$.

Step 7: Find the length of the line segment RS. The length of the line segment $RS$ is the absolute difference between the x-coordinates of $R$ and $S$: $|RS| = \left| \left(1 + \frac{2\sqrt{3}}{3}\right) - \left(1 - \frac{2\sqrt{3}}{3}\right) \right| = \left| \frac{4\sqrt{3}}{3} \right| = \frac{4\sqrt{3}}{3}$

Common Mistakes & Tips

• Sign Errors: Be very careful with signs when separating variables and integrating.
• Completing the Square: Ensure you correctly complete the square to obtain the standard form of the circle's equation.
• Normal to a Circle: Remember that the normal to a circle always passes through its center. This significantly simplifies the problem.

Summary

We solved the given differential equation to find the equation of a circle. Using the initial condition and the given area, we determined the circle's center and radius. We then found the points of intersection of the circle with the y-axis, determined the equations of the normals at these points, and calculated the distance between their x-intercepts. The length of the line segment $RS$ is $\frac{4\sqrt{3}}{3}$.

Final Answer The final answer is $\boxed{\frac{4\sqrt{3}}{3}}$, which corresponds to option (A).