Q. 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 $(1/6)^{\text{th}}$ of the perimeter of $C$, then $\beta-\sqrt3\alpha$ is equal to
A) $3+\sqrt3$
B) $4$
C) $4-\sqrt3$
D) $3$
Explanation
[Solution by Dr. Abubacker] The given circle has centre $(1,-2)$ and radius $3$. Reflecting the centre in $2x-3y+5=0$ (value $2+6+5=13$): $(1,-2)-\frac{2\cdot13}{13}(2,-3)=(-3,4)$, so $C$ has centre $O(-3,4)$, radius $3$. Then $A=(0,4)$. An arc equal to $1/6$ of the perimeter subtends $60^\circ$ at $O$; as $\beta<4$, $B=\left(-3+3\cos60^\circ,\,4-3\sin60^\circ\right)=\left(-\frac32,\,4-\frac{3\sqrt3}2\right)$. Hence $\beta-\sqrt3\alpha=4-\frac{3\sqrt3}2+\frac{3\sqrt3}2=4$. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.