The maximum value of y=(x−3)2+x2−22−x2+x2−12 is

y=f(x)=x2−22+(x−3)2−x2+x2−12 Note that the first term describes the distance between Px,x2 and A(3,2) where as the second term describes the  distance between Px,x2 and B(0,1) .  Now PA−PB≤AB for possible positions of P Hencef(x)max= distance between AB=9+1=10