Let circle C$C$ be the image of x^2 + y^2 - 2x + 4y - 4 = 0$x^{2} + y^{2} - 2x + 4y - 4 = 0$ in the line 2x - 3y + 5 = 0$2x - 3y + 5 = 0$ and A$A$ be the point on C$C$ such that OA$OA$ is \parallel to the x-axis and A$A$ lies on the \right hand side of the centre O$O$ of C$C$. If B(alpha,beta)$B(\alpha,\beta)$, with beta < 4$\beta < 4$, lies on C$C$ such that the length of the arc AB$AB$ is (1/6)^textth$(1/6)^{\text{th}}$ of the perimeter of C$C$, then beta - sqrt3alpha$\beta - \sqrt{3}\alpha$ is equal to :

### Related Formula The coordinates for the reflection image of a point (x_1, y_1)$(x_1, y_1)$ across a standard line ax + by + c = 0$ax + by + c = 0$ are determined using: fracx - x_1a = fracy - y_1b = frac-2(ax_1 + by_1 + c)a^2 + b^2$$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2}$$ ### Core Logic Find the center and radius of the original given circle: x^2 + y^2 - 2x + 4y - 4 = 0 implies textCenter = (1, -2), \, r = sqrt1^2 + (-2)^2 - (-4) = 3$$x^2 + y^2 - 2x + 4y - 4 = 0 \implies \text{Center} = (1, -2), \, r = \sqrt{1^2 + (-2)^2 - (-4)} = 3$$

Reflect the center point (1, -2)$(1, -2)$ across the line mirror 2x - 3y + 5 = 0$2x - 3y + 5 = 0$: fracx - 12 = fracy + 2-3 = frac-2(2(1) - 3(-2) + 5)2^2 + (-3)^2 = frac-2(2 + 6 + 5)13 = -2$$\frac{x - 1}{2} = \frac{y + 2}{-3} = \frac{-2(2(1) - 3(-2) + 5)}{2^2 + (-3)^2} = \frac{-2(2 + 6 + 5)}{13} = -2$$ x - 1 = -4 implies x = -3$$x - 1 = -4 \implies x = -3$$ y + 2 = 6 implies y = 4$$y + 2 = 6 \implies y = 4$$ Thus, the center O$O$ of the reflected circle C$C$ is (-3, 4)$(-3, 4)$, and its radius is preserved at r = 3$r = 3$. ### Step 1: Locate Point A We are given that OA$OA$ is \parallel to the x-axis, meaning its y-coordinate matches the center. Since A$A$ lies to the \right of the center O(-3, 4)$O(-3, 4)$: A = (-3 + r, \, 4) = (-3 + 3, \, 4) = (0, 4)$$A = (-3 + r, \, 4) = (-3 + 3, \, 4) = (0, 4)$$ ### Step 2: Determine Angular Position of Point B The arc length AB$AB$ is given as frac16$\frac{1}{6}$ of the total perimeter: textArc length = rtheta = frac16(2pi r) implies theta = fracpi3 = 60^circ$$\text{Arc length} = r\theta = \frac{1}{6}(2\pi r) \implies \theta = \frac{\pi}{3} = 60^\circ$$

Using parametric coordinates relative to center O(-3, 4)$O(-3, 4)$ with radius r = 3$r = 3$: alpha = -3 + 3costheta, quad beta = 4 + 3sintheta$$\alpha = -3 + 3\cos\theta, \quad \beta = 4 + 3\sin\theta$$ Since beta < 4$\beta < 4$, the \angle theta$\theta$ must point downwards into the negative quadrant relative to A$A$, meaning theta = -60^circ = -fracpi3$\theta = -60^\circ = -\frac{\pi}{3}$: alpha = -3 + 3cosleft(-fracpi3right) = -3 + 3left(frac12 ight) = -frac32$$\alpha = -3 + 3\cos\left(-\frac{\pi}{3}\right) = -3 + 3\left(\frac{1}{2} ight) = -\frac{3}{2}$$ beta = 4 + 3sinleft(-fracpi3right) = 4 - frac3sqrt32$$\beta = 4 + 3\sin\left(-\frac{\pi}{3}\right) = 4 - \frac{3\sqrt{3}}{2}$$ ### Step 3: Evaluate Final Algebraic Value Substitute the determined coordinates into the target expression: beta - sqrt3alpha = left(4 - frac3sqrt32right) - sqrt3left(-frac32 ight)$$\beta - \sqrt{3}\alpha = \left(4 - \frac{3\sqrt{3}}{2}\right) - \sqrt{3}\left(-\frac{3}{2} ight)$$ beta - sqrt3alpha = 4 - frac3sqrt32 + frac3sqrt32 = 4$$\beta - \sqrt{3}\alpha = 4 - \frac{3\sqrt{3}}{2} + \frac{3\sqrt{3}}{2} = 4$$ ### Pattern Recognition Whenever parametric configurations on a circle involve radical coordinate multipliers like beta - sqrt3alpha$\beta - \sqrt{3}\alpha$, using angular vectors centered at the origin of the circle avoids setting up and solving long distance equations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Circles