两圆x*x+y*y+2kx+k*k-1=0与x*x+y*y+2(k+1)y+k*k+2k=0的圆心之间最短距离是多少?

两圆x*x+y*y+2kx+k*k-1=0与x*x+y*y+2(k+1)y+k*k+2k=0的圆心之间最短距离是多少?
两圆x*x+y*y+2kx+k*k-1=0与x*x+y*y+2(k+1)y+k*k+2k=0的圆心之间最短距离是多少?

两圆x*x+y*y+2kx+k*k-1=0与x*x+y*y+2(k+1)y+k*k+2k=0的圆心之间最短距离是多少?
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答案是1/2
两圆方程化简为：
(x+k)^2+y^2=1
x^2+(y+k+1)^2=1
可知两圆的圆心分别为：(-k,0),(0,-k-1)
距离d^2=k^2+(k+1)^2=2k^2+2k+1=2(k^2+k+1/4+1/4)=2[(k+1/2)^2+1/4]
所以，当k=-1/2时，d有最小值dmin=1/2.

半径：
R=1，r=1
圆心：（-k,0),(0,-k-1)
距离d
=√[k^2+(k+1)^2
=√[2(k+1/2)^2+1/2]
k=-1/2
dmin=√2/2