Question

1.已知圆的一条直径的端点分别是A（x1,y1) ,B (x2,y2),求证此圆的方程是

Answer 1

1.

圆心（(x1+x2)/2,(y1+y2)/2)

半径^2=(1/4)AB^2=(1/4)((x1-x2)^2+(y1-y2)^2)

此圆的方程:

(x-(x1+x2)/2)^2+(y-(y1+y2)/2)^2=(1/4)((x1-x2)^2+(y1-y2)^2)

(x-x1)(x-x2)+(y-y1)(y-y2)=0

2.

设C(x,y)

AC^2=AB^2

(x-4)^2+(y-2)^2=(4-3)^2+(2-5)^2

(x-4)^2+(y-2)^2=10

是圆,但要把点（3,5)及点（5,-1)去除