一般情况下的阿波罗尼斯圆的方程的推导过程
首先介绍一下什么是阿波罗尼斯圆:
已知平面上两点 A,BA, BA,B, 则所有满足 PAPB=k\frac{PA}{PB}=kPBPA=k 且不等于 111 的点 PPP 的轨迹是一个以定比 m:nm:nm:n 内分和外分定线段 ABABAB 的两个分点的连线为直径的圆. 这个轨迹最先由古希腊数学家阿波罗尼斯发现, 故称作阿氏圆.
我翻了半天知乎, 发现没有人写阿波罗尼斯圆在最一般的情况下方程的推导, 遂作此文. 当然, 很大一部分原因是我下边写的这种推导十分麻烦, 不如以所给的两定点为 xxx 轴来研究.
那么下边我将以最一般的情况, 用解析几何的方法来推导.
在平面直角坐标系中，已知 AAA 的坐标为 (x1,y1)(x_1,y_1)(x1,y1)， BBB 的坐标为 (x2,y2)(x_2,y_2)(x2,y2) （ A,BA,BA,B 互异），设动点 PPP 的坐标为 (x,y)(x,y)(x,y), 满足 ∣PA∣∣PB∣=k(k>0且k≠1)\frac{|{PA}|}{|{PB}|}=k \quad (k>0 且 k\neq1)∣PB∣∣PA∣=k(k>0且k=1). 易得
∣PA∣=(x−x1)2+(y−y1)2,∣PB∣=(x−x2)2+(y−y2)2. \begin{gather*} |PA| = \sqrt{(x-x_1)^2+(y-y_1)^2} ,\\ |PB| = \sqrt{(x-x_2)^2+(y-y_2)^2} . \end{gather*} ∣PA∣=(x−x1)2+(y−y1)2,∣PB∣=(x−x2)2+(y−y2)2.
由 ∣PA∣∣PB∣=k\frac{|{PA}|}{|{PB}|}=k∣PB∣∣PA∣=k, 即 ∣PA∣=k∣PB∣|PA| = k|PB|∣PA∣=k∣PB∣, 对两边平方后将上一行的两式带入得
(x−x1)2+(y−y1)2=k2[(x−x2)2+(y−y2)2], (x-x_1)^2+(y-y_1)^2=k^2[(x-x_2)^2+(y-y_2)^2] , (x−x1)2+(y−y1)2=k2[(x−x2)2+(y−y2)2],
展开整理后得到
(1−k2)x2+(1−k2)y2+2(k2x2−x1)x+2(k2y2−y1)y+x12−k2x22+y12−k2y22=0, (1-k^2)x^2+(1-k^2)y^2+2(k^2x_2-x_1)x+2(k^2y_2-y_1)y+x_1^2-k^2x_2^2+y_1^2-k^2y_2^2=0 , (1−k2)x2+(1−k2)y2+2(k2x2−x1)x+2(k2y2−y1)y+x12−k2x22+y12−k2y22=0,
等号两边同除以 (1−k2)(1-k^2)(1−k2), 得到
x2+y2+2(k2x2−x1)1−k2x+2(k2y2−y1)1−k2y+x12−k2x22+y12−k2y221−k2=0. x^2+y^2+\frac{2(k^2x_2-x_1)}{1-k^2}x+\frac{2(k^2y_2-y_1)}{1-k^2}y+\frac{x_1^2-k^2x_2^2+y_1^2-k^2y_2^2}{1-k^2}=0 . x2+y2+1−k22(k2x2−x1)x+1−k22(k2y2−y1)y+1−k2x12−k2x22+y12−k2y22=0.
可以配一下方，那么可以得到
(x+k2x2−x11−k2)2+(y+k2y2−y11−k2)2= (k2x2−x1)2+(k2y2−y1)2+(k2−1)(x12−k2x22+y12−k2y22)(1−k2)2= k4x22−2k2x1x2+x12+k4y22−2k2y1y2+y12+(k2−1)x12+(1−k2)k2x22+(k2−1)y12+(1−k2)k2y22(1−k2)2= k2x12+k2x22−2k2x1x2+k2y12+k2y22−2k2y1y2(1−k2)2= k2(1−k2)2[(x1−x2)2+(y1−y2)2],\begin{equation}\notag \begin{split} &\ \left(x+\frac{k^2x_2-x_1}{1-k^2}\right)^2+\left(y+\frac{k^2y_2-y_1}{1-k^2}\right)^2 \\ =&\ \frac{(k^2x_{2}-x_{1})^2+(k^2y_{2}-y_{1})^2+(k^2-1)(x_{1}^2-k^2x_{2}^2+y_{1}^2-k^2y_{2}^2)}{(1-k^2)^2} \\ =&\ \frac{k^4x_{2}^2-2k^2x_{1}x_{2}+x_{1}^2+k^4y_{2}^2-2k^2y_{1}y_{2}+y_{1}^2+(k^2-1)x_{1}^2+(1-k^2)k^2x_{2}^2+(k^2-1)y_{1}^2+(1-k^2)k^2y_{2}^2}{(1-k^2)^2} \\ =&\ \frac{k^2 x_{1}^2+k^2x_{2}^2-2 k^2x_{1} x_{2}+k^2 y_{1}^2+k^2y_{2}^2-2k^2y_{1}y_{2}}{(1-k^2)^2} \\ =&\ \frac{k^2}{(1-k^2)^2}[(x_{1}-x_{2})^2+(y_{1}-y_{2})^2] ,\\ \end{split} \end{equation}==== (x+1−k2k2x2−x1)2+(y+1−k2k2y2−y1)2 (1−k2)2(k2x2−x1)2+(k2y2−y1)2+(k2−1)(x12−k2x22+y12−k2y22) (1−k2)2k4x22−2k2x1x2+x12+k4y22−2k2y1y2+y12+(k2−1)x12+(1−k2)k2x22+(k2−1)y12+(1−k2)k2y22 (1−k2)2k2x12+k2x22−2k2x1x2+k2y12+k2y22−2k2y1y2 (1−k2)2k2[(x1−x2)2+(y1−y2)2],
即
(x+k2x2−x11−k2)2+(y+k2y2−y11−k2)2=k2(1−k2)2[(x1−x2)2+(y1−y2)2]. (x+\frac{k^2x_{2}-x_{1}}{1-k^2})^2+(y+\frac{k^2y_{2}-y_{1}}{1-k^2})^2=\frac{k^2}{(1-k^2)^2}[(x_{1}-x_{2})^2+(y_{1}-y_{2})^2] . (x+1−k2k2x2−x1)2+(y+1−k2k2y2−y1)2=(1−k2)2k2[(x1−x2)2+(y1−y2)2].
由于 k>0k>0k>0 且 k≠1k\neq 1k=1, 并且 A,BA,BA,B 两点互异, 所以等号右边的式子始终大于 000 , 所以这便是圆的标准方程.
这样我们就得到了阿波罗尼斯圆的方程. 实际上, 也就证明了阿波罗尼斯圆的命题.
由上式可知此情况下圆心为
(k2x2−x11−k2,k2y2−y11−k2). \left(\frac{k^2x_{2}-x_{1}}{1-k^2},\frac{k^2y_{2}-y_{1}}{1-k^2}\right) . (1−k2k2x2−x1,1−k2k2y2−y1).
观察最后得到的这个式子, 不难发现
(x1−x2)2+(y1−y2)2=∣AB∣2. (x_1-x_2)^2+(y_1-y_2)^2=|AB|^2 . (x1−x2)2+(y1−y2)2=∣AB∣2.
那么, 阿波罗尼斯圆的半径可以表示为
R=∣k1−k2∣⋅∣AB∣. R=\left|\frac{k}{1-k^2}\right|\cdot|AB| . R=1−k2k⋅∣AB∣.
综上
在平面直角坐标系中, 已知 AAA 的坐标为 (x1,y1)(x_1,y_1)(x1,y1), BBB 的坐标为 (x2,y2)(x_2,y_2)(x2,y2) (A,BA,BA,B 互异), 设动点 PPP 的坐标为 (x,y)(x,y)(x,y), 满足 ∣PA∣∣PB∣=k(k>0且k≠1)\frac{|PA|}{|PB|}=k \quad (k>0且k\neq 1)∣PB∣∣PA∣=k(k>0且k=1), 那么点 PPP 的轨迹是阿波罗尼斯圆, 方程为
(x+k2x2−x11−k2)2+(y+k2y2−y11−k2)2=(k1−k2∣AB∣)2, \left(x+\frac{k^2x_{2}-x_{1}}{1-k^2}\right)^2+\left(y+\frac{k^2y_{2}-y_{1}}{1-k^2}\right)^2=\left(\frac{k}{1-k^2}|AB|\right)^2 , (x+1−k2k2x2−x1)2+(y+1−k2k2y2−y1)2=(1−k2k∣AB∣)2,
圆心为
(k2x2−x11−k2,k2y2−y11−k2), \left(\frac{k^2x_{2}-x_{1}}{1-k^2},\frac{k^2y_{2}-y_{1}}{1-k^2}\right) , (1−k2k2x2−x1,1−k2k2y2−y1),
半径为
∣k1−k2∣⋅∣AB∣. \left|\frac{k}{1-k^2}\right|\cdot|AB| . 1−k2k⋅∣AB∣.
以上.