Question

已知P为原x^2+y^2=16上一动点,点A(12,0)当P在圆上运动时,求线段中点轨迹方程.

Answer 1

设B（x,y）为线段中点,p(x1,y1)

x=1/2(x1+12) y=1/2(y1+0)

得 x1=2x-12 y1=2y

又x1^2+y1^2=16

所以(2x-12)^2+(2y)^2=16

即(x-6)^2+y^2=4