Question

点P是圆x2+y2-8x-2y+13=0上的动点,O是坐标原点,则线段OP的中点Q的轨迹方程

Answer 1

C:(x-4)^2+(y-1)^2=4

Q(x,y)

P(2x,2y)

(xP-4)^2+(yP-1)^2=4

(2x-4)^2+(2y-1)^2=4

(x-2)^2+(y-0.5)^2=1