两圆x2+y2+2x-4y+3=0与x2+y2-4x+2y+3=0上的两点的最短距离是?

Question

如题所述

Answer 1

x2+y2+2x-4y+3=0
(x+1)^2+(y-2)^2=2 以点(-1,2)为圆心,√2为半径
x2+y2-4x+2y+3=0
(x-2)^2+(y+1)^2=2 以点(2,-1)为圆心,√2为半径
两圆圆心距为：√[(-1-2)^2+(2+1)^2=3√2>2√2
所以最短距离为：3√2-2√2=√2