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:
Step-by-Step Solution
Key Concept: Find the centre of the circle by completing the square: $(x-2)^2+(y-8)^2=4$, centre $C=(2,8)$. The normal to $y^2=4x$ at point $(m^2,-2m)$ (parametric form) passes through $C=(2,8)$. Normal: $y=mx-2m-m^3$. Substitute $(2,8)$: $8=2m-2m-m^3\Rightarrow m=-2$. Point on parabola: $P=(4,4)$.
Centre $C=(2,8)$. Normal at $P$: $m=-2\Rightarrow P=(4,4)$. $d^2=(4-2)^2+(4-8)^2=4+16=20$.
Correct Answer: 3