JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Circle

Writing the circle as $x^2+y^2+2gx+2fy+c=0$ gives $g=\frac{p}{2}$, $f=\frac{1-p}{2}$, $c=5$, so

The condition $0<r\le 5$ means $0<r^2\le 25$, i.e. $0<2p^2-2p-19\le 100$. The first inequality gives $p<\frac{1-\sqrt{39}}{2}$ or $p>\frac{1+\sqrt{39}}{2}$, and the second gives $\frac{1-\sqrt{239}}{2}\le p\le \frac{1+\sqrt{239}}{2}$. Over the admissible $p$, the quantity $q=p^2$ takes every value in $\left(\frac{20-\sqrt{39}}{2},\frac{120+\sqrt{239}}{2}\right]\approx (6.88,67.73]$. The integers here are $7,8,\dots ,67$, so the set has $61$ elements.