已知圆x²;+y²+x-6y+m=0和直线x+2y-3=0相交于A,B两点,且OA,⊥OB（O为坐标

由题知,
已知圆x²+y²+x-6y+m=0和直线x+2y-3=0相交于A,B两点,
令两式联立得
x²+y²+x-6y+m=0
x+2y-3=0
即
1.25x²+2.5x+m-6.75=0
设A(x1,y1),B(x2,y2)
所以,
x1+x2 = -2
x1*x2 = 0.8m-5.4
因为OA⊥OB
所以,
x1x2+y1y2
=x1x2+(3-x1)(3-x2)/4
=1.25x1x2-0.75(x1+x2)+2.25
=1.25(0.8m-5.4)-0.75(-2)+2.25
=0
解得
m=3