JEE Main 2025 — 24 January, Morning Shift — Question 15

$\text{Original\; circle:\; }x^{2}+y^{2}-2x+4y-4=0.$

$\text{Complete\; the\; square:\; }(x-1)^2+(y+2)^2=9\Rightarrow \text{centre\; }(1,-2),\ r=3.$

$\text{Reflect\; the\; centre\; }(1,-2)\text{ in\; the\; line\; }2x-3y+5=0.$

$\text{With\; }a=2,\ b=-3,\ c=5,\ d=\frac{ax_0+by_0+c}{a^2+b^2}=\frac{2\cdot1+(-3)(-2)+5}{4+9}=\frac{13}{13}=1.$

$\text{Reflection:\; }O=(x',y')=(x_0-2ad,\;y_0-2bd)=(1-4,\;-2-2(-3))=(-3,4).$

$\text{So }O=(-3,4),\ r=3.$

$\text{Point\; }A\; \text{ to\; the\; right\; of\; }O\text{ with\; }OA\parallel x\text{-axis is }A=O+(r,0)=(0,4).$

$\text{Arc\; }AB\; \text{ equals\; } \tfrac{1}{6}\text{ of\; circumference\; }\Rightarrow$ central angle $=\tfrac{\pi}{3}=60^\circ.$

$\text{Since\; }B\; \text{ has\; } \beta<4,\ B\text{ is\; the\; clockwise\; point\; at\; angle\; }-60^\circ:$

$B=O+r(\cos(-60^\circ),\sin(-60^\circ))=(-3,4)+3\Big(\tfrac{1}{2},-\tfrac{\sqrt3}{2}\Big)=\Big(-\tfrac{3}{2},\,4-\tfrac{3\sqrt3}{2}\Big).$

$\therefore \alpha=-\tfrac{3}{2},\ \beta=4-\tfrac{3\sqrt3}{2}.$

$\beta-\sqrt3\,\alpha=\Big(4-\tfrac{3\sqrt3}{2}\Big)-\sqrt3\Big(-\tfrac{3}{2}\Big)=4.$

$\boxed{4}$