Question

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为坐标原点）求M的值

Answer 1

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