JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Circle

Rewriting the circle equation: $x^2+y^2-3x+2y+\frac{k}{4}=0$

Center: $\left(\frac{3}{2},-1\right)$, Radius: $r=\sqrt{\frac{9}{4}+1-\frac{k}{4}}=\frac{\sqrt{13-k}}{2}$

For the circle to exist: $k\le 13$.

Condition 1: Point $\left(1,-\frac{1}{3}\right)$ lies on or inside the circle:
$1+\frac{1}{9}-3-\frac{2}{3}+\frac{k}{4}\le 0$
$\frac{k}{4}\le \frac{23}{9}$
$k\le \frac{92}{9}$

Condition 2: Circle lies in fourth quadrant means center is in fourth quadrant and circle doesn't touch axes. The center $\left(\frac{3}{2},-1\right)$ is in the fourth quadrant. For the circle to not touch the $x$-axis: $r<1$, so $\frac{\sqrt{13-k}}{2}<1$, giving $k>9$.

Combining all conditions: $k\in \left(9,\frac{92}{9}\right]$

Answer: (D)