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If the shortest distance of the parabola y^2=4x from the centre of the circle x^2+y^2-4x-16y+64=0 is d, then d^2 is equal to :

JEE Main 2024 — Mathematics Coordinate Geometry

2024mcqmedium

If the shortest distance of the parabola y2=4x{y}^{2}=4xy2=4x from the centre of the circle x2+y2−4x−16y+64=0{x}^{2}+{y}^{2}-4x-16y+64=0x2+y2−4x−16y+64=0 is ddd, then d2{d}^{2}d2 is equal to :

Official previous-year question

Held on 27 Jan 2024 · Verified 6 Jul 2026.

Options

  1. A

    161616

  2. B

    242424

  3. C

    202020

  4. D

    363636

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