If the shortest distance of the parabola y2=4x from the centre of the circle x2+y2−4x−16y+64=0 is d, then d2 is equal to:
Solutions · 1
Official solution
Circle centre (2,8). Point on parabola: (t2,2t). Minimize
\[g(t)=(t^2-2)^2+(2t-8)^2,\qquad g'(t)=4t(t^2-2)+4(2t-8)=4t^3-32=0 \Rightarrow t=2.\]
Point (4,4): d2=(4−2)2+(4−8)2=4+16=20.