JEE Main 2025 Jan 24 Shift 1, Mathematics Q15: Introduction to Circles

The original circle has centre $(1,-2)$ and radius $\sqrt{1+4+4}=3$. Reflect the centre in $2x-3y+5=0$. With $d=2(1)-3(-2)+5=13$ and $a^2+b^2=13$, the reflected point is

$(1,-2)-\frac{2\cdot 13}{13}(2,-3)=(1-4,-2+6)=(-3,4)$.

So circle $C$: $(x+3{)}^2+(y-4{)}^2=9$, centre $O(-3,4)$, radius $3$.

Point $A$ on $C$ with $OA$ parallel to $x$-axis, to the right of $O$: $A=(-3+3,4)=(0,4)$.

Arc length $=\frac{1}{6}$ of perimeter means subtended angle $\theta =\frac{2\pi }{6}=\frac{\pi }{3}$.

$B$ is on the circle at angle $\frac{\pi }{3}$ from $A$ as seen from $O$. With $\beta <4$, take $B$ below $A$:

$B=O+3(\cos (-\pi /3),\sin (-\pi /3))\cdot$ (relative to the direction $OA$, which is along $+x$)

Position vector $OA=(3,0)$; rotating by $-\pi /3$:

$\overrightarrow{OB}=(3\cos (\pi /3),-3\sin (\pi /3))=\left(\frac{3}{2},-\frac{3\sqrt{3}}{2}\right)$

$B=O+\overrightarrow{OB}=\left(-3+\frac{3}{2},4-\frac{3\sqrt{3}}{2}\right)=\left(-\frac{3}{2},4-\frac{3\sqrt{3}}{2}\right)$

So $\alpha =-\frac{3}{2}$, $\beta =4-\frac{3\sqrt{3}}{2}$.

$\beta -\sqrt{3}\alpha =4-\frac{3\sqrt{3}}{2}-\sqrt{3}\cdot \left(-\frac{3}{2}\right)=4-\frac{3\sqrt{3}}{2}+\frac{3\sqrt{3}}{2}=4$.

$\therefore$ Option (4).