JEE Main 2025 Jan 22 Shift 2, Mathematics Q19: Involving Two or More Circles

Let $z=x+iy$. Then $z(1+i)+\bar{z}(1-i)=(x+iy)(1+i)+(x-iy)(1-i)$. The imaginary parts cancel and the real part is $2x-2y$, so the locus is $2x-2y=4\Rightarrow x-y=2$.

The region $|z-3|\le 1$ is the closed disk centred at $(3,0)$ with radius 1.

Distance from centre $(3,0)$ to line $x-y=2$: $d=\frac{|3-0-2|}{\sqrt{2}}=\frac{1}{\sqrt{2}}$.

The line cuts the disk into a smaller segment (closer to a point of the disk away from the centre) and a larger segment. The chord subtends an angle $2\theta$ at the centre where $\cos \theta =\frac{d}{r}=\frac{1}{\sqrt{2}}$, so $\theta =\frac{\pi }{4}$ and the central angle is $\frac{\pi }{2}$.

Area of the smaller segment: $A_{\text{small}}=\frac{1}{2}{r}^2(2\theta -\sin 2\theta )=\frac{1}{2}\left(\frac{\pi }{2}-1\right)=\frac{\pi }{4}-\frac{1}{2}$.

Area of disk: $\pi$. Larger segment: $\pi -\left(\frac{\pi }{4}-\frac{1}{2}\right)=\frac{3\pi }{4}+\frac{1}{2}$.

$|\alpha -\beta |=\left(\frac{3\pi }{4}+\frac{1}{2}\right)-\left(\frac{\pi }{4}-\frac{1}{2}\right)=\frac{\pi }{2}+1$.