两圆(x-1)^2+(y-1)^2=2和(x+2)^2+(y-4)^2=3的公切线的条数是

问题描述:

两圆(x-1)^2+(y-1)^2=2和(x+2)^2+(y-4)^2=3的公切线的条数是

(x-1)^2+(y-1)^2=2 圆心(1,1) 半径r1=根号2
(x+2)^2+(y-4)^2=3 圆心(-2,4) 半径r2=根号3
圆心距=3倍根号2 =4.2几 半径和为=根号2+根号3=3.1几
两圆相离 故有四条公切线