已知圆x2+y2+x-6y+3=0上的两点P,Q关于直线kx-y+4=0对称,且OP⊥OQ(O为坐标原

已知圆x2+y2+x-6y+3=0上的两点P,Q关于直线kx-y+4=0对称,OPOQ(O为坐标原点),则直线PQ的方程为      

答案

 y=-x+y=-x+

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