JEE Advanced 2021 — Paper 1 — Question 28

Rewrite the circle: $x^2+y^2+5x-3y+4=0$ as $(x+\frac{5}{2})^2+(y-\frac{3}{2})^2 = \frac{9}{2}$. So center $C = -\frac{5}{2}+\frac{3}{2}i$, radius $R = \frac{3}{\sqrt{2}}$. The condition $\arg\left(\frac{z+\alpha}{z+\beta}\right)=\frac{\pi}{4}$ describes a circular arc with endpoints $-\alpha,-\beta$. The inscribed angle is $\frac{\pi}{4}$, so the central angle subtended by chord $(-\alpha,-\beta)$ is $\frac{\pi}{2}$. For the whole set of $z$ to lie on the given circle, that circle must be exactly this arc's circle. Hence its center is $C$ and radius $R$. Step 4: The endpoints $-\alpha,-\beta$ lie on the circle. Their coordinates (real axis) satisfy $x^2+5x+4=0$, giving $x=-1,-4$. So $\{\alpha,\beta\}=\{1,4\}$. The chord length is $|\alpha-\beta| = 3$. This equals $R\sqrt{2} = \frac{3}{\sqrt{2}}\cdot\sqrt{2}=3$, confirming consistency. The midpoint of the chord is $M = \left(-\frac{\alpha+\beta}{2},0\right)$. Distance from $C$ to $M$ must be $R/\sqrt{2} = \frac{3}{2}$: $\left(\frac{\alpha+\beta-5}{2}\right)^2 + \left(\frac{3}{2}\right)^2 = \left(\frac{3}{2}\right)^2$, so $\alpha+\beta=5$. Both $(1,4)$ and $(4,1)$ have sum 5, but only $\alpha=1,\beta=4$ makes $\arg\left(\frac{z+\alpha}{z+\beta}\right)=\frac{\pi}{4}$ for points on the upper arc (e.g., test a point). Thus $\alpha=1,\beta=4$, giving $\alpha\beta=4$ and $\beta=4$. The true statements are B and D.