过点A（4,0）作直线L交圆O：x^2+y^2=4于B,C两点,求线段BC中点P的轨迹方程.

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设B(x1,y1),C(x2,y2),BC中点为M(x0,y0),则
x1²+y1²=4,
x2²+y2²=4
以上两式相减,得(x1+x2)(x1-x2)+(y1-y2)(y1+y2)=0
又x0=(x1+x2)/2,y0=(y1+y2)/2
∴(y1-y2)/(x1-x2)=-(x1+x2)/(y1+y2)=-x0/y0,即k=-x0/y0.
又点M(x0,y0)和电A(4,0)也在直线L上,k=y0/(x0-4)
∴-x0/y0=y0/(x0-4),整理得 x0²-4x0+y0²=0
所以,点M的轨迹方程为x²-4x+y²=0,即(x-2)²+y²=4.