Question

已知圆x^2+y^2=4上一点A(2,0),B(1,1)为圆内一点,P,Q为圆上的动点.

Answer 1

(1）求线段AP中点的轨迹方程

AP中点(x,y)

xP=2x-2,yP=2y

x^2+y^2=4

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

AP中点的轨迹方程:(x-1)^2+y^2=1

(2）若角PBQ=90°,求PQ中点的轨迹方程

PQ中点(x,y)

xP+xQ=2x,(xP+xQ)^2=(2x)^2

(xP)^2+(xQ)^2+2xP*xQ=4x^2.(1)

yP+yQ=2y

(yP)^2+(yQ)^2+2yP*yQ=4y^2.(2)

(xP)^2+(yP)^2=4.(3)

(xQ)^2+(yQ)^2=4.(4)

角PBQ=90°

k(PB)*k(QB)=-1

[(yP-1)/(xP-1)]*[(yQ-1)/(xQ-1)]=-1

xP*xQ+yP*yQ=(xP+xQ)+(yP+yQ)-2=2x+2y-2

2xP*xQ+2yP*yQ=4x+4y-4.(5)

(1)+(2)-(3)-(4)-(5):

x^2+y^2-x-y-1=0

PQ中点的轨迹方程圆:

(x-0.5)^2+(y-0.5)^2=1.5