Question

已知关于x的方程C:x^2+y^2-2x-4y+m=0,问:

Answer 1

y=x+2代入圆得：

x^2+x^2+4x+4-2x-4x-8+m=0

x^2-x-2+m/2=0,根为x1,x2,则有：x1+x2=1,x1x2=-2+m/2

交点为M(x1,x1+2),N(x2,x2+2)

OM⊥ON,得：(x1+2)/x1*(x2+2)/x2=-1

即:x1x2+2(x1+x2)+4=-x1x2

即x1x2+(x1+x2)+2=0

即：-2+m/2+1+2=0

得：m=-2