Question

过圆x^2+y^2=4 内一点A(1,0)作圆的弦,求这些弦的中点M的轨迹方程

Answer 1

设弦AB坐标分别是:A(X1,Y1),B(X2,Y2),中点M(X,Y)

则有:X1+X2=2X,Y1+Y2=2Y

X1^2+Y1^2=4

X2^2+Y2^2=4

二式相减得:

(X1+X2)(X1-X2)+(Y1+Y2)(Y1-Y2)=0

(Y1-Y2)/(X1-X2)=-2X/2Y=-X/Y

即AB的斜率是:-X/Y

又K(AB)=K(MA)=(Y-0)/(X-1)

所以,-X/Y=Y/(X-1)

即方程是:Y^2=-X(X-1)