x^2+y^2-2x-4y+m=0
(x-1)²+(y-2)²=5-m
圆心到直线的距离:
l=|1+2*2-4|/√(1²+2²)
5-m=l²+l² 构成一个等腰直角三角形
m=3
M(x1,y1) N(x2,y2)
OM⊥ON
y1/x1*y2/x2=-1
y1y2=-x1x2
x^2+y^2-2x-4y+m=0
x+2y-4=0
(4-2y)²+y^2-2(4-2y)-4y+m=0
5y²+8-16y+m=0
y1y2=(8+m)/5
y1+y2=16/5
x1x2
=(4-2y1)(4-2y2)
=16-8y2-8y1+4y1y2
=(4m-16)/5
y1y2=-x1x2
(8+m)/5=-(4m-16)/5
8+m=16-4m
5m=8
m=8/5