JEE 2024 — Mathematics PYQ

$x^2+y^2-4x-16y+64=0$

$\therefore$ Centre $(2,8)$ and radius $=2$

Normal $y+tx=2t+t^2$

The normal will pass through $(2,8)$

$\Rightarrow 8+2t=2t+t^3$

$\Rightarrow t^3=8\Rightarrow t=2$

$\therefore p(t)=p(t^2,2t)=p(4,4)$

$\therefore$ Shortest distance $=\sqrt{(2-4{)}^2+(8-4{)}^2}$

$=\sqrt{4+16}=\sqrt{20}\Rightarrow d^2=20$