Question

已知圆x²＋y²＋x－6y＋m＝0与直线x＋2y－3＝0相交于P·Q两点,O为坐标原点,若OP⊥OQ,求实数m

Answer 1

x^2+y^2+x-6y+m=0与x+2y-3=0相交于P、Q两点

x=3-2y

x^2+y^2+x-6y+m=0

(3-2y)^2^2+y^2+(3-2y)-6y+m=0

5y^2-20y+12+m=0

yP+yQ=4

yP*yQ=(12+m)/5

xP=3-2yP,xQ=3-2yQ

xP*xQ=(3-2yP)*(3-2yQ)=9-6(yP+yQ)+4yP*yQ

OP⊥OQ

k(OP)*k(OQ)=-1

(yP/xP)*(yQ/xQ)=-1

xP*xQ+yP*yQ=0

[9-6(yP+yQ)+4yP*yQ]+yP*yQ=0

9-6(yP+yQ)+5yP*yQ=0

9-6*4+5*(12+m)/5=0

m=3