Key Concepts and Formulas

• Equation of a Circle: The standard form of a circle with center $(h,k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.
• Reflection of a Point about a Line: The image $(x',y')$ of a point $(x_1,y_1)$ reflected across the line $ax+by+c=0$ is given by the formula: $\frac{x'-x_1}{a} = \frac{y'-y_1}{b} = \frac{-2(ax_1+by_1+c)}{a^2+b^2}$
• Arc Length Formula: The length $s$ of an arc of a circle with radius $r$ subtending an angle $\theta$ (in radians) at the center is $s = r\theta$. The perimeter of a circle is $2\pi r$.
• Parametric Form of a Circle: A point $(x,y)$ on a circle with center $(h,k)$ and radius $r$ can be represented as $(h+r\cos\phi, k+r\sin\phi)$, where $\phi$ is the angle made by the radius connecting the center to the point, with respect to the positive x-axis.

Step-by-Step Solution

1. Determine the Center and Radius of the Initial Circle

The given equation of the initial circle is $x^2+y^2-2 x+4 y-4=0$. We want to rewrite this in standard form to identify the center and radius. Grouping the $x$ and $y$ terms, we have: $(x^2-2x) + (y^2+4y) = 4$ Completing the square for the $x$ terms: $(x^2-2x+1) - 1 = (x-1)^2 - 1$. Completing the square for the $y$ terms: $(y^2+4y+4) - 4 = (y+2)^2 - 4$. Substituting these back into the equation: $(x-1)^2 - 1 + (y+2)^2 - 4 = 4$ $(x-1)^2 + (y+2)^2 = 9$

• Explanation: Completing the square allows us to rewrite the equation in standard form, making it easy to identify the center and radius.
• Result: The initial circle has center $O_1(1,-2)$ and radius $r = \sqrt{9} = 3$.

2. Find the Center of Circle $C$ by Reflection

Circle $C$ is the image of the initial circle in the line $2x-3y+5=0$. The radius remains the same upon reflection, but the center changes. We need to find the reflection of $O_1(1,-2)$ across the line $2x-3y+5=0$. Using the reflection formula for a point $(x_1,y_1)=(1,-2)$ about the line $ax+by+c=0$ where $a=2, b=-3, c=5$: $\frac{x'-x_1}{a} = \frac{y'-y_1}{b} = \frac{-2(ax_1+by_1+c)}{a^2+b^2}$ Substitute the values: $\frac{x'-1}{2} = \frac{y'-(-2)}{-3} = \frac{-2(2(1)-3(-2)+5)}{2^2+(-3)^2}$ $\frac{x'-1}{2} = \frac{y'+2}{-3} = \frac{-2(2+6+5)}{4+9}$ $\frac{x'-1}{2} = \frac{y'+2}{-3} = \frac{-2(13)}{13}$ $\frac{x'-1}{2} = \frac{y'+2}{-3} = -2$ Now, solve for $x'$ and $y'$: $x'-1 = 2(-2) \Rightarrow x'-1 = -4 \Rightarrow x' = -3$ $y'+2 = -3(-2) \Rightarrow y'+2 = 6 \Rightarrow y' = 4$

• Explanation: This formula directly provides the coordinates of the reflected point. We apply it to the center of the original circle to find the center of circle $C$.
• Result: The center of circle $C$ is $O(-3,4)$. Its radius is still $r=3$.

3. Determine the Coordinates of Point $A$

Point $A$ is on circle $C$ such that $OA$ is parallel to the x-axis and $A$ lies to the right of the center $O$. Since $OA$ is parallel to the x-axis, the y-coordinate of $A$ is the same as the y-coordinate of $O$, which is 4. Let $A = (x_A, 4)$. The equation of circle $C$ is $(x+3)^2 + (y-4)^2 = 9$. Substituting $y=4$, we get: $(x+3)^2 + (4-4)^2 = 9$ $(x+3)^2 = 9$ $x+3 = \pm 3$ $x = -3 \pm 3$ So, $x = 0$ or $x = -6$. Since $A$ lies to the right of the center $O(-3,4)$, we choose the larger x-coordinate, which is $x=0$. Thus, $A = (0,4)$.

• Explanation: We use the given information about the position of A relative to the center O to uniquely determine its coordinates.
• Result: The coordinates of point $A$ are $(0,4)$.

4. Determine the Coordinates of Point $B$

The length of the arc $AB$ is $\frac{1}{6}$ of the perimeter of $C$. The perimeter of $C$ is $2\pi r = 2\pi(3) = 6\pi$. Thus, the arc length $AB$ is $\frac{1}{6}(6\pi) = \pi$. Using the arc length formula $s = r\theta$, we have $\pi = 3\theta$, so $\theta = \frac{\pi}{3}$. Let $B = (\alpha, \beta)$. We know that the center of the circle is $O(-3,4)$. Let $\phi_A$ be the angle that $OA$ makes with the positive x-axis, and $\phi_B$ be the angle that $OB$ makes with the positive x-axis. Since $A=(0,4)$, and $O=(-3,4)$, $OA$ is along the positive x-axis relative to O, so $\phi_A = 0$. The angle between $OA$ and $OB$ is $\frac{\pi}{3}$. Thus, $\phi_B = \phi_A + \frac{\pi}{3} = 0 + \frac{\pi}{3} = \frac{\pi}{3}$. Using the parametric form of the circle, we have: $\alpha = -3 + 3\cos\left(\frac{\pi}{3}\right) = -3 + 3\left(\frac{1}{2}\right) = -3 + \frac{3}{2} = -\frac{3}{2}$ $\beta = 4 + 3\sin\left(\frac{\pi}{3}\right) = 4 + 3\left(\frac{\sqrt{3}}{2}\right) = 4 + \frac{3\sqrt{3}}{2}$ Since $\beta < 4$, we must consider the other possible angle. The angle between OA and OB can also be $2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$. In this case, $\alpha = -3 + 3\cos\left(\frac{5\pi}{3}\right) = -3 + 3\left(\frac{1}{2}\right) = -3 + \frac{3}{2} = -\frac{3}{2}$ $\beta = 4 + 3\sin\left(\frac{5\pi}{3}\right) = 4 + 3\left(-\frac{\sqrt{3}}{2}\right) = 4 - \frac{3\sqrt{3}}{2}$ Since $\beta < 4$, this is the correct value for $\beta$.

• Explanation: We use the arc length to find the angle subtended at the center, and then use the parametric form to find the coordinates of point B. We consider both possible angles since the arc length is specified, but the direction is not.
• Result: $B = \left(-\frac{3}{2}, 4 - \frac{3\sqrt{3}}{2}\right)$

5. Calculate $\beta - \sqrt{3}\alpha$

We have $\alpha = -\frac{3}{2}$ and $\beta = 4 - \frac{3\sqrt{3}}{2}$. Then $\beta - \sqrt{3}\alpha = \left(4 - \frac{3\sqrt{3}}{2}\right) - \sqrt{3}\left(-\frac{3}{2}\right) = 4 - \frac{3\sqrt{3}}{2} + \frac{3\sqrt{3}}{2} = 4$

• Explanation: We substitute the values of $\alpha$ and $\beta$ into the expression to obtain the final result.
• Result: $\beta - \sqrt{3}\alpha = 4$.

Common Mistakes & Tips

• Reflection Formula Sign Errors: Be very careful with the signs in the reflection formula. A small error here will propagate through the rest of the solution.
• Arc Length Ambiguity: When using the arc length to determine the position of point B, remember that there are two possible locations for B (clockwise or counterclockwise from A). The condition $\beta < 4$ is crucial for selecting the correct position.
• Completing the Square: Double-check your work when completing the square to avoid mistakes in determining the center and radius of the circle.

Summary

We first found the center and radius of the initial circle. Then, we reflected the center across the given line to find the center of circle $C$, keeping the radius the same. Next, we used the information about point $A$ to determine its coordinates. Finally, we used the arc length to find the coordinates of point $B$ and computed $\beta - \sqrt{3}\alpha$, which equals 4.

Final Answer

The final answer is \boxed{4}, which corresponds to option (C).