Question

已知点A(2,0),B(4,0)和圆(x-3)^2+(y-4)^2=4上动点p,求|PA|^2+|PB|^2的最小值是P

Answer 1

设P(x,y)，满足(x-3)^2+(y-4)^2=4

圆心为C(3,4),半径r=2

那么|PA|^2+|PB|^2

=(x-2)^2+y^2+(x-4)^2+y^2

=2x^2+2y^2-12x+20

=2[(x-3)^2+y^2]+2

令D(3,0)

所以|PD|=√[(x-3)^2+y^2]

|PD|min=|CD|-r=2