Question

过一动点分别向两相交圆引切线,切线长相等,则动点的轨迹是

Answer 1

设两圆为方程

(x-x1)^2+(y-y1)^2=r1^2 圆心O1为(x1,y1),半径为r1

(x-x2)^2+(y-y2)^2=r2^2 圆心O2为(x1,y1),半径为r1

动点P(x,y)

勾股定理得到切线长相等

PO1 ^2-r1^2 =PO2 ^2-r2^2

(x-x1)^2+(y-y1)^2-r1^2 =(x-x2)^2+(y-y2)^2-r2^2

x^2和y^2系数为0,所以动点的轨迹应该是直线