Question

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O

Answer 1

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