Let circle C' be the image of C: x^2+y^2-2x+4y-4=0 in the line 2x-3y+5=0.

Question

Let circle $C'$ be the image of $C: x^2+y^2-2x+4y-4=0$ in the line $2x-3y+5=0$. Let $A$ be the point on $C$ such that $OA$ is parallel to the $x$-axis and lies on the right-hand side of the centre of $C$. If $B(\alpha,\beta)$, with $\beta<4$, lies on $C'$ such that the length of the arc $AB$ is $\dfrac16$th of the perimeter of $C'$, then $\beta-\sqrt3\alpha$ is equal to:

Accepted Answer

$C$: centre $(1,-2)$, radius $3$. Reflecting $(1,-2)$ in $2x-3y+5=0$ gives the centre of $C'$ at $(-3,4)$, so $C':(x+3)^2+(y-4)^2=9$.

*(The remaining steps locating $A$ and $B$ precisely and confirming $\beta-\sqrt3\alpha=\dfrac43$ were too corrupted in the source to reconstruct reliably beyond this point; the given answer is option C.)*

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