Let circle $C$ be the image of $x^2 + y^2 - 2x + 4y - 4 = 0$ in the line $2x - 3y + 5 = 0$ and $A$ be the point on $C$ such that $OA$ is parallel to $x$-axis and $A$ lies on the right hand side of the centre $O$ of $C$. If $B(\alpha, \beta)$, with $\beta < 4$, lies on $C$ such that the length of the arc $AB$ is $\tfrac{1}{6}^{\text{th}}$ of the perimeter of $C$, then $\beta - \sqrt{3}\,\alpha$ is equal to
Step-by-Step Solution
Key Concept: Reflect the centre of the original circle across the given line to locate $O$; arc $= \tfrac{1}{6} \cdot 2\pi r$ implies central angle $\tfrac{\pi}{3}$; rotate $\overrightarrow{OA}$ by $-\tfrac{\pi}{3}$ (clockwise, since $\beta < 4$) to find $B$.
Original circle: centre $(1,-2)$, $r=3$. Reflecting $(1,-2)$ in $2x-3y+5=0$: the signed distance parameter is $-2(2\cdot1+(-3)(-2)+5)/13=-2$, giving $O=(-3,4)$. Circle $C$: $(x+3)^2+(y-4)^2=9$. $A$ is rightmost point on the horizontal through $O$: $A=(0,4)$. Arc $AB=\tfrac{1}{6}\cdot2\pi(3)=\pi$, so central angle $\theta=\tfrac{\pi}{3}$. Since $\beta<4$ rotate $\overrightarrow{OA}=(1,0)$ by $-\tfrac{\pi}{3}$: $$B=(-3,4)+3\left(\tfrac{1}{2},-\tfrac{\sqrt{3}}{2}\right)=\left(-\tfrac{3}{2},\ 4-\tfrac{3\sqrt{3}}{2}\right).$$ Therefore $\beta-\sqrt{3}\,\alpha=\left(4-\tfrac{3\sqrt{3}}{2}\right)-\sqrt{3}\left(-\tfrac{3}{2}\right)=4-\tfrac{3\sqrt{3}}{2}+\tfrac{3\sqrt{3}}{2}=4$.
Correct Answer: 2