Question

圆x^2+y^2-2ax+a^2=2和圆x^2+y^2-2by+b^2=1外离,则a,b满足的条件是?为什么答案是大于3

Answer 1

x^2+y^2-2ax+a^2=2可化为(x-a)^2+y^2=2

x^2+y^2-2by+b^2=1可化为x^2+(y-b)^2=1

两圆外离,得 圆心距＞半径之和

即 a^2+b^2＞(根号2+1)^2=3+2根号2