圆:x^2+(y+m)^2=4与圆:(x-m)^2+y^2=36有4条公切线,则实数m的范围是?
问题描述：
圆:x^2+(y+m)^2=4与圆:(x-m)^2+y^2=36有4条公切线,则实数m的范围是?
答
圆:x^2+(y+m)^2=4与圆:(x-m)^2+y^2=36有4条公切线,
则两圆相离
即圆心距大于半径之和
圆心距=√(2m^2)
半径之和=2+6=8
√(2m^2)>8
m^2>32
m>4√2或m