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?

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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?
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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?
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?

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?
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

呵呵,哥哥,你耍我呢,这个是什么公式啊,爱因斯坦他老人家也没发明出来啊

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<-4√