JEE Main 202427 Jan, Morning shiftMathematicsEasyCoordinate Geometry478× since 2002If the shortest distance of the parabola y2=4xy^2=4 xy2=4x from the centre of the circle x2+y2−4x−16y+64=0x^2+y^2-4 x-16 y+64=0x2+y2−4x−16y+64=0 is d\mathrm{d}d, then d2\mathrm{d}^2d2 is equal to :AOption A. 16BOption B. 24COption C. 20DOption D. 36CheckSkip · show solution
