Question

设圆x^2+y^2+x-6y+m=0的圆心为C,它与直线x-y-2=0交于P,Q两点,若CP⊥CQ,求m

Answer 1

由 x-y-2=0
得 y=x-2
x^2+(x-2)^2+x-6(x-2)+m=0
x^2+x^2-4x+4+x-6x+12+m=0
2x^2-9x+16+m=0
∴ x1+x2=9/2 x1*x2=(16+m)/2
y1+y2=x1-2+x2-2=x1+x2-4=9/2-4=1/2
y1*y2=(x1-2)*(x2-2)=x1*x2-2(x1+x2)+4
=(16+m)/2-2*9/2+4=(16+m)/2-5
=(6+m)/2
设 P,Q两点为P(x1,y1),Q(x2,y2)
求 圆x^2+y^2+x-6y+m=0的圆心C
x^2+y^2+x-6y+m=x^2+x+1/4+y^2-6y+9-1/4-9+m
=(x+1/2)^2+(y-3)^2+m-1/4-9
所以 C(-1/2,3)
∵ CP⊥CQ
∴ (y1-3)/(x1+1/2)*(y2-3)/(x2+1/2)=-1
(y1-3)*(y2-3)=-(x1+1/2)*(x2+1/2)
y1*y2-3(y1+y2)+9=-[x1*x2+1/2*(x1+x2)+1/4]
(6+m)/2-3*1/2+9=-[(16+m)/2+1/2*9/2+1/4]
(6+m)/2+15/2=-[(16+m)/2+5/2]
m+21=-m-21
2m=-2*21
∴ m=-21