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

Centre $(1,-2), r=3$ Reflection of $(1,-2)$ about $2 x-3 y+5=0$$\begin{aligned} & \frac{x-1}{2}=\frac{y+2}{-3}=\frac{-2(2+6+5)}{13}=-2 \\ & x=-3, y=4 \end{aligned}$Equation of circle ' C ' $C:(x+3)^2+(y-4)^2=9$ A.T.Q.<img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/ERL7MBpRPnzfcTf_afHfMdiy0bLJLjyUh3LHiuqCofc.original.fullsize.png"><br>\begin{aligned} & \ell(\operatorname{arcAB})=\frac{1}{6} \times 2 \pi \mathrm{r} \\ & \mathrm{r} \theta=\frac{1}{6} \times 2 \pi \mathrm{r} \\ & \theta=\frac{\pi}{3} \\ & (\alpha+6)^2+(\beta-4)^2=27 \\ & \frac{(\alpha+3)^2 \pm(\beta-4)^2=9}{(\alpha+6)^2-(\alpha+3)^2=18} \\ & \Rightarrow 6 \alpha=-9\end{aligned}$$\begin{aligned} & \Rightarrow \alpha=\frac{-3}{2}, \beta=\left(4-\frac{3 \sqrt{3}}{2}\right) \\ & \therefore \beta-\sqrt{3} \alpha \\ & \left(4-\frac{3 \sqrt{3}}{2}\right)+\frac{3 \sqrt{3}}{2} \\ & =4\end{aligned}$