JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Parabola

The parabola $y^2=4x$ has focus $(1,0)$. A general point on it is $(t^2,2t)$, and its distance from $(a,0)$ satisfies


Differentiating with respect to $t$ and setting to zero:


This gives $t=0$ (distance $a$) or $t^2=a-2$ (requires $a>2$), with


Since the shortest distance is 4, and $a>2$ is needed for a nonzero-$t$ critical point that beats $t=0$:


The circle passes through $(5,0)$ and the focus $(1,0)$, both on the x-axis, with its centre also on the x-axis. The centre must be equidistant from both points, so it lies on their perpendicular bisector — the vertical line $x=3$ — and on the x-axis, giving centre $(3,0)$ and radius $2$.