Question

已知圆x^2+y^2+x-6y+m=0与直线x+2y-3=0相交于P、Q两点，O为坐标原点，

Answer 1

圆x^2+y^2+x-6y+m=0与直线x+2y-3=0 ∴X=3-2Y代入前式得：5y²-20y+12+m=0 ∴y1+y2=4 x1+x2=6-2(y1+y2)=-2 y1*y2=(12+m)/5 x1*x2=-15+4(12+m)/5

PQ=√1+(-1/2)²|X1-X2|=√5/2*√(X1+X2)²-4X1X2=13 ∴m=41/4

y1/x1*(y2/x2)=-1 ∴[(12+m)/5 ]/ [ -15+4(12+m)/5]=-1 ∴m=-254/32