JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Circle

The parabola $y=x^2-6x+12=(x-3{)}^2+3$ has vertex $P(3,3)$. The circle $x^2+y^2-2x-4y+3=0$ has centre $C(1,2)$ and radius $\sqrt{1+4-3}=\sqrt{2}$, with $PC=\sqrt{(3-1{)}^2+(3-2{)}^2}=\sqrt{5}>\sqrt{2}$, so $P$ lies outside it. If a line through $P$ meets the circle at $R,S$ and $M$ is the midpoint of chord $RS$, then with $d$ the distance from $C$ to the line, $PM=\sqrt{P{C}^2-d^2}$ while the half-chord is $\sqrt{r^2-d^2}$, so $PR+PS=2PM=2\sqrt{P{C}^2-d^2}$. This is greatest when $d=0$, i.e. the line passes through $C$, giving $(PR+PS{)}_{\max }=2PC=2\sqrt{5}$. Therefore the maximum value of $(PR+PS{)}^2$ is $20$.