解:圆心M(0,2),AB中点G(r,s),切点(x,y),Q(m,0)
x^2+(y-2)^2=1............(1)
MQ^2=MB^2+BQ^2
m^2+4=1+(x-m)^2+y^2
=4y-2mx-3+x^2+(y-2)^2=4y-2mx-2
整理:mx-2y+3=0..............(2)
(1),(2)连立:
(4+m^2)x^2-2mx-3=0
r=(x1+x2)/2=m/(m^2+4)........(3)
(4+m^2)y^2-4(3+m^2)y+3m^3+9=0
s=2(3+m^2)/(m^2+4)............(4)
(3),(4)连立消掉参数m:
r^2+s^2-7s/2+3=0
所以AB中点轨迹方程:
x^2+y^2-7y/2+3=0
追问为什么 MQ^2=MB^2+BQ^2
追答 QB是圆M的切线
三角形MBQ是直角三角形
所以MQ^2=MB^2+BQ^2
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