已知圆x2+y2+x-6y+m=0与直线x+2y-3=0相交于P、Q两点,O为原点,且OP⊥OQ,求实数m的值．

已知圆x2+y2+x-6y+m=0与直线x+2y-3=0相交于P、Q两点,O为原点,且OP⊥OQ,求实数m的值．由x+2y-3=0得x=3-2y代入x2+y2+x-6y+m=0化简得：5y2-20y+12+m=0y1+y2=4,y1•y2= (12+m)/5设P、Q的坐标分别为（x1,y1）、（x2,y2）,由OP⊥OQ可得：x1•x2+y1•y2=0x1•x2+y1•y2=（3-2y1）•（3-2y2）+y1•y2=9-6（y1+y2）+5y1•y2=9-6×4+5× (12+m)/5=m-3=0解得：m=3由OP⊥OQ可得：x1•x2+y1•y2=0这个为啥 快