Question

已知点P（3,4）为圆C：x^2+y^2=64内一点,圆周上有两动点A、B,当∠APB= 90°时,以AP、BP为邻边,

Answer 1

设A(x1,y1) B(x2,y2) AB中点M(x,y),应有

x=x1+x2/2 ==>x1+x2=2x①

y=y1+y2/2 ==>y1+y2=2y②

又AB两点都在圆上,有

x1^2+y1^2=64③

x2^2+y2^2=64④

* =0

(x1-3)(x2-3)+(y1-4)(y2-4)=0

x1x2+y1y2-3(x1+x2)-4(y1+y2)+25=0⑤

又由③+④得（x1^2+x2^2）+(y1^2+y2^2)=128

(x1+x2)^2+(y1+y2)^2-2(x1x2+y1y2)=128

代入①②得4x^2+4y^2-2(x1x2+y1y2)=128

x1x2+y1y2=2x^2+2y^2-64⑥

⑥代入⑤2x^2+2y^2-64-6x-8y+25=0即M点轨迹为

2x^2+2y^2-6x-8y-39=0

显然M点为Q点和P点的中点,设Q（m,n）于是有

x=m+3/2

y=n+4/2

代入M的方程2(m+3/2)^2+2(n+4/2)^2-6(m+3/2)-8(n+4/2)-39=0

我不化简了