两圆x^2+y^2-6y=0与x^2+y^2-8y+12=0的位置关系是?
人气：418 ℃ 时间：2020-05-17 16:14:13
解答
x^2+y^2-6y=0
化简 x^2+(y-3)^2=9
圆心（0,3）半径 3
x^2+y^2-8y+12=0
化简 x^2+(y-4)^2=4
圆心（0,4）半径 2
所以是内切关系
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