> 数学 >
圆x2+y2+2kx+k2-1=0与圆x2+y2+2(k+1)y+k2+2k=0的圆心之间的最短距离是(  )
A.
2
2

B. 2
2

C. 1
D.
2
人气:359 ℃ 时间:2020-04-15 02:24:47
解答
圆x2+y2+2kx+k2-1=0的圆心(-k,0),圆x2+y2+2(k+1)y+k2+2k=0的圆心(0,-k-1).
圆心之间的距离为:
(−k)2+(k+1)2
=
2k2+2k+1
=
2(k+
1
2
)
2
+
1
2

当k=
1
2
时圆心距最小,最小值为:
2
2

故选:A.
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