已知圆C:x²+y²+x-6y+m=0与直线l:x+2y-3=0相交于P,Q两点,O为原点,若向量OP·向量OQ=0,
问题描述：
已知圆C:x²+y²+x-6y+m=0与直线l:x+2y-3=0相交于P,Q两点,O为原点,若向量OP·向量OQ=0,
求实数m的大小.
请用圆系方程来回答,
答
P(x1,y1),Q(x2,y2)
圆C:x^2+y^2+x-6y+m=0 (1)
直线l:x+2y-3=0 (2)
sub (2) into (1)
(3-2y)^2 +y^2 +(3-2y)-6y +m =0
5y^2-14y+12+m=0
y1y2 = (12+m)/5
Similarly,
x^2+[(3-x)/2]^2+x-6[(3-x)/2]+m=0
4x^2+(3-x)^2+4x-12(3-x)+4m=0
5x^2+10x+4m-27=0
y1y2 = (4m-27)/5
OP.OQ=0
x1x2+y1y2=0
(12+m)/5 + (4m-27)/5 =0
12+m+4m-27=0
m =5正解为m=3不好意思，算错！OP.OQ=0x1x2+y1y2=0(12+m)/5 + (4m-27)/5 =012+m+4m-27=05m =15m =3