设圆x2+y2-4x-5=0的弦AB的中点为P(3,1),则直线AB的方程为(  ) A.x+y-4=0 B.x+y-5=0 C.x-y+4=0 D.x-y+5=0

问题描述:

设圆x2+y2-4x-5=0的弦AB的中点为P(3,1),则直线AB的方程为(  )
A. x+y-4=0
B. x+y-5=0
C. x-y+4=0
D. x-y+5=0

圆x2+y2-4x-5=0的圆心O(2,0),
圆x2+y2-4x-5=0的弦AB的中点为P(3,1),
∴直线AB与直线OP垂直,
∵kOP=

1−0
3−2
=1,∴kAB=-1,
∴直线AB的方程为:y-1=-(x-3),整理,得:x+y-4=0.
故选:A.