猜想有两解,试试.
①求圆心坐标O（x0,y0).
分析：OA=OB,且AO与BO的夹角为k°.
依题意有 (x1-x0)²+(y1-y0)²=(x2-x0)²+(y2-y0)²,①
(x1-x2)²+(y1-y2)²=2[(x1-x0)²+(y1-y0)²]-2[(x1-x0)²+(y1-y0)²]cosk°.②
即x1²+y1²-2x1x0-2y1y0=x2²+y2²-2x2x0-2y2y0,③
(x1-x2)²+(y1-y2)²=2[(x1-x0)²+(y1-y0)²](1-cosk°).④
化简得2(x2-x1)x0+2(y2-y1)y0+(x1-x2)²+(y1-y2)²-x2²-y2²=0,⑤
(x1-x0)²+(y1-y0)²=[(x1-x2)²+(y1-y2)²]/[2(1-cosk°)].⑥
由⑤得x0=[x2²+y2²-(x1-x2)²-(y1-y2)²-2(y2-y1)y0]/[2(x2-x1)]
=[x1x2+y1y2-x1²-y1²-2(y2-y1)y0]/[2(x2-x1)]
=[x1(x2-x1)+y1(y2-y1)-2(y2-y1)y0]/[2(x2-x1)]
=[x1(x2-x1)+(y1-2y0)(y2-y1)]/[2(x2-x1)]
=x1/2+(y1-2y0)(y2-y1)/[2(x2-x1)],⑦
把⑦代入⑥得
{x1/2-(y1-2y0)(y2-y1)/[2(x2-x1)]}²+(y1-y0)²=[(x1-x2)²+(y1-y2)²]/[2(1-cosk°)],
x1²/4+(y1-2y0)²(y2-y1)²/[4(x2-x1)²]-x1(y1-2y0)(y2-y1)/[2(x2-x1)]+(y1-y0)²=[(x1-x2)²+(y1-y2)²]/[2(1-cosk°)],
稍后.