求两圆x^2+y^2-2x-6y+9=0与x^2+y^2-2mx-2(m-1)y+2m^2-2m=0的圆心距的最小值,并判断当这个圆心距取得最小值时两圆的位置关系

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求两圆x^2+y^2-2x-6y+9=0与x^2+y^2-2mx-2(m-1)y+2m^2-2m=0的圆心距的最小值,并判断当这个圆心距取得最小值时两圆的位置关系
求两圆x^2+y^2-2x-6y+9=0与x^2+y^2-2mx-2(m-1)y+2m^2-2m=0的圆心距的最小值,并判断当这个圆心距取得最小值时两圆的位置关系

求两圆x^2+y^2-2x-6y+9=0与x^2+y^2-2mx-2(m-1)y+2m^2-2m=0的圆心距的最小值,并判断当这个圆心距取得最小值时两圆的位置关系
x^2+y^2-2x-6y+9=0
(x-1)^2+(y-3)^2=1
圆心(1,3)
x^2+y^2-2mx-2(m-1)y+2m^2-2m=0
(x-m)^2+(y-m+1)^2=1
圆心(m ,m- 1)
圆心距^2
=(m-1)^2+(m-1-3)^2
=2m^2-10m +17
=2(m-2.5)^2+4.5
m=2.5,最小值3√2 /2
3√2 /2>1+1
两圆的位置关系：外离

x^2+y^2-2x-6y+9=0 => (x-1)^2+(y-3)^2 = 1
x^2+y^2-2mx-2(m-1)y+2m^2-2m=0 => (x-m)^2+(y-(m-1))^2 = 1
所以圆心距离 d^2 = (m-1)^2 + (m-1-3)^2 = 2m^2 - 10m + 17
当m = 10/4=5/2的时候取得最小值 d^2 = 25/2 - 50 + 17 = 9/2
d = 3/√2 > 2
所以此时两圆相离

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