圆1：x^2+y^2-4x+4y+6=0 -> (x-2)^2+(y+2)^2=2 ->圆心(2,-2),r1=√2
圆2：x^2+y^2-6x+2y+1=0 -> (x-3)^2+(y+1)^2=9 ->圆心(3,-1),r2=3
根据两点距离公式得d=√[(x1-x2)^2+(y1-y2)^2]=√2
因为d<|r2-r1|
所以两圆关系为包含