三点坐标求圆方程

1. 问题

求过A(x1,y1),B(x2,y2),C(x3,y3)A\left(x_1, y_1\right), B\left(x_2, y_2\right), C\left(x_3, y_3\right)三点的圆方程MM.

2. 求解

2.1. 解方程法

设圆心M(x0,y0)M(x_0,y_0), 设线段ABAB的中点为PP, 设线段ACAC的中点为QQ, 则根据垂径定理, 有

可得方程组

①(x0−xP)(x2−x1)+(y0−yP)(y2−y1)=0\red{① \quad (x_0-x_P)(x_2-x_1)+(y_0-y_P)(y_2-y_1)=0}

②(x0−xQ)(x3−x1)+(y0−yQ)(y3−y1)=0②\quad (x_0-x_Q)(x_3-x_1)+(y_0-y_Q)(y_3-y_1)=0

第②②式可变形为

③(x0−xP)(x3−x1)+(y0−yP)(y3−y1)=t\red{③\quad (x_0-x_P)(x_3-x_1)+(y_0-y_P)(y_3-y_1)=t}

其中, t=(xQ−xP)(x3−x1)+(yQ−yP)(y3−yp)t=(x_Q-x_P)(x_3-x_1)+(y_Q-y_P)(y_3-y_p). 接下来就是加减消元.

由 (y2−y1)×③−(y3−y1)×①(y_2-y_1)\times③-(y_3-y_1)\times① 得

x0−xP=t(y2−y1)(x3−x1)(y2−y1)−(x2−x1)(y3−y1)x_0-x_P=\dfrac{t(y_2-y_1)}{(x_3-x_1)(y_2-y_1)-(x_2-x_1)(y_3-y_1)}.

由 (x2−x1)×③−(x3−x1)×①(x_2-x_1)\times③-(x_3-x_1)\times① 得

y0−yP=t(x2−x1)(y3−y1)(x2−x1)−(y2−y1)(x3−x1)y_0-y_P=\dfrac{t(x_2-x_1)}{(y_3-y_1)(x_2-x_1)-(y_2-y_1)(x_3-x_1)}.

解出圆心x0,y0x_0,y_0后, 半径r=∣MA→∣r=|\overrightarrow{MA}|.

如果用中垂线交点列方程那就慢了!

这里解方程组是有技巧的, 那就是先变形再消元, 这种技巧非常实用!

2.2. 待定系数法(推荐)

以BCBC为直径的圆方程:

F1(x,y):=(x−x2)(x−x3)+(y−y2)(y−y3)=0F_1(x, y):=\left(x-x_2\right)\left(x-x_3\right)+\left(y-y_2\right)\left(y-y_3\right)=0

设P(x,y)P(x,y)在BCBC为直径的圆上, 则PB→⋅PC→=0\overrightarrow{PB}\cdot\overrightarrow{PC}=0, 整理就是上式.

直线BCBC方程:

F2(x,y):=(x−x2)(y3−y2)−(y−y2)(x3−x2)=0F_2(x, y):=\left(x-x_2\right)\left(y_3-y_2\right)-\left(y-y_2\right)\left(x_3-x_2\right)=0

设Q(x,y)Q(x,y)在直线BCBC上, 取BC→\overrightarrow{BC}的一个法向量(即与BC→\overrightarrow{BC}垂直的向量)
n=((y3−y2),  −(x3−x2))\bm n=\Big((y_3-y_2),\; - \left(x_3-x_2\right)\Big),
则根据关系式QB→⋅n=0\overrightarrow{QB}\cdot \bm{n}=0, 整理可得上式

过 B,CB, C两点的圆可以表示为

F1(x,y)+λF2(x,y)=0F_1(x, y)+\lambda F_2(x, y)=0

只需验证两条(它就是所谓的圆系方程):
(1)B,CB,C两点坐标满足上面等式成立;
(2)上式可整理成圆的标准式
第一条容易验证, 第二条略微麻烦. 这里给出第二条验证的思路, 不妨假设BCBC的中点为坐标原点(否则做平移处理), 则从形式上看, 有
F1(x,y)=x2+y2−r2F_1(x,y)=x^2+y^2-r^2
F2(x,y)=2ax+2byF_2(x,y)=2ax+2by
即F1(x,y)+λF2(x,y)=(x+λa)2+(y+λb)2−r2−λ2(a2+b2)=0F_1(x, y)+\lambda F_2(x, y)=(x+\lambda a)^2+(y+\lambda b)^2-r^2-\lambda^2(a^2+b^2)=0 为圆方程

把AA点坐标代入上式, 可解出,

λ=−F1(x1,y1)F2(x1,y1)\lambda=-\dfrac{F_1(x_1,y_1)}{F_2(x_1,y_1)},

因此过A,B,CA,B,C三点的圆方程为:

F1(x,y)−F1(x1,y1)F2(x1,y1)F2(x,y)=0F_1(x, y)-\dfrac{F_1(x_1,y_1)}{F_2(x_1,y_1)} F_2(x, y)=0

2.3. 四点共圆复数法

熟知结论:

设P(x,y)P(x,y)在圆上, 设复数

z1=(x2−x1)+(y2−y1)i(x3−x1)+(y3−y1)iz_1=\dfrac{(x_2-x_1)+ (y_2-y_1)\mathrm{i}}{(x_3-x_1)+ (y_3-y_1)\mathrm{i}}, z2=(x2−x)+(y2−y)i(x3−x)+(y3−y)iz_2=\dfrac{(x_2-x)+ (y_2-y)\mathrm{i}}{(x_3-x)+ (y_3-y)\mathrm{i}},

则根据复数乘法几何意义, z1z_1的辐角对应于角∠BAC\angle BAC, z2z_2的辐角对应于角∠BPC\angle BPC. 由于P,A,B,CP,A,B,C四点共圆, 因此复数z1z_1与z2z_2的辐角相等或互补, 于是, 根据第三条结论可得圆MM方程.

这里给出相关表达式

  • k1=1/[(x3−x1)2+(y3−y1)2]k_1=1/[(x_3-x_1)^2+(y_3-y_1)^2]
  • k2=1/[(x3−x)2+(y3−y)2]k_2=1/[(x_3-x)^2+(y_3-y)^2]

它们和待定系数法里的公式对应关系为:

2.4. 三元一次方程组

假设圆方程为: x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0

于是有

{x12+y12+2gx1+2fy1+c=0x22+y22+2gx2+2fy2+c=0x32+y32+2gx3+2fy3+c=0\begin{cases} x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c = 0\\ x_2^2 + y_2^2 + 2gx_2 + 2fy_2 + c = 0\\ x_3^2 + y_3^2 + 2gx_3 + 2fy_3 + c = 0 \end{cases}

最后解出参数g,f,cg,f,c

2.5. 复数推导(推荐)

设圆心为 (x0,x0)(x_0,x_0), 构造复数

c   =(x0−x3)+i(y0−y3)c\;\,=(x_0-x_3)+i(y_0-y_3)
z1=(x1−x3)+i(y1−y3)z_1=(x_1-x_3)+i(y_1-y_3)
z2=(x2−x3)+i(y2−y3)z_2=(x_2-x_3)+i(y_2-y_3)

则r2=ccˉ=(c−z1)c−z1‾=(z−z2)c−z2‾r^2=c\bar{c}=(c-z_1)\overline{c-{z}_1}=(z-z_2)\overline{c-{z}_2}

根据中间两项

ccˉ=ccˉ+z1zˉ1−czˉ1−z1cˉc\bar{c}=c\bar{c}+z_1\bar{z}_1-c\bar{z}_1-z_1\bar{c}

有

czˉ1+cˉz1=z1zˉ1c\bar{z}_1+\bar{c}z_1=z_1\bar{z}_1

同理

czˉ2+cˉz2=z2zˉ2c\bar{z}_2+\bar{c}z_2=z_2\bar{z}_2

解得

c=z2z1zˉ1−z1z2zˉ2z2zˉ1−z1zˉ2c=\dfrac{z_2z_1\bar{z}_1- z_1z_2\bar{z}_2}{z_2\bar{z}_1-z_1\bar{z}_2}

最后回代得

这种方法结论很简洁, 可直接得到圆心坐标

3. 总结

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