已知直线L:(m+1)x+(m-1)y-3m-1=0 圆C:x^2+y^2=9 (1)判定直线L与圆的位置关系 (2)若L与C交于AB两点 求当AB

已知直线L:(m+1)x+(m-1)y-3m-1=0 圆C:x^2+y^2=9 (1)判定直线L与圆的位置关系 (2)若L与C交于AB两点 求当AB最小值时直线L的方程 (求过程答案) 谢谢

(m+1)x+(m-1)y-3m-1=0
(m+1)x+(m-1)y=3m+1
(x+y)m+x-y=3m+1
x+y=3
x-y=1
则x=2,y=1
直线L:(m+1)x+(m-1)y-3m-1=0恒过(2,1)
圆C:x^2+y^2=9
则直线与圆C相交

设AB的中点为N
当圆C:x^2+y^2=9 的圆心O与N的连线ON⊥AB时,AB最小
则N点即(2,1)
向量ON;(2,1)
直线的方向向量就是(-1,2)
直线方程为2x+y-5=0
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第1个回答  2011-09-25
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第2个回答  2011-09-25
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第3个回答  2011-10-02
分析:圆心(0,0)半径3.和直线L的方程可得d,d与r的关心进行分类讨论,L交AB,则L为圆C直径