三点确定圆心算法推导

已知a,b,c三点求过这三点的圆心坐标

$a(x_1,y_1)$ 、$b(x_2,y_2)$ 、$c(x_3,y_3)$

确认三点是否共线

叉积计算方式
$$\overrightarrow{v}(X_1,Y_1)\times \overrightarrow{u}(X_2,Y_2)=X_1{Y}_2-X_2{Y}_1$$
使用叉积判断三个点是否共线
$$\begin{gathered} \overrightarrow{ab}=((x_2-x_1),(y_2-y_1)) \\ \overrightarrow{ac}=((x_3-x_1),(y_3-y_1)) \end{gathered}$$
带入上方可到的：
$$\begin{array}{rl}value& =(x_2-x_1)(y_3-y_1)-(x_3-x_1)(y_2-y_1)\\ value& =x_2{y}_3-x_2{y}_1-x_1{y}_3+x_1{y}_1-x_3{y}_2+x_3{y}_1+x_1{y}_2-x_1{y}_1\end{array}$$
整合后得到
$$value=x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$$
$$\{\begin{array}{llc}value& <0& 点b在\overrightarrow{ac}右边\\ value& >0& 点b在\overrightarrow{ac}左边\\ value& =0& 三点共线\end{array}$$

不共线的三点才可以确认一个圆

三点求解圆心

已知圆的方程为
$$(x-x_0{)}^2+(y-y_0{)}^2=r^2$$

假设圆心坐标为$O(a,b)$带入三点可以得到下列方程
$$\{\begin{array}{ll}(x_1-a{)}^2+(y_1-b{)}^2=r^2& ①\\ (x_2-a{)}^2+(y_2-b{)}^2=r^2& ②\\ (x_3-a{)}^2+(y_3-b{)}^2=r^2& ③\end{array}$$
拆解后可得
$$\{\begin{array}{ll}{x}_1^2-2x_{1}a+a^2+y_1^2-2y_{1}b+b^2=r^2& ①\\ x_2^2-2x_{2}a+a^2+y_2^2-2y_{2}b+b^2=r^2& ②\\ x_3^2-2x_{3}a+a^2+y_3^2-2y_{3}b+b^2=r^2& ③\end{array}$$
使用①-②，②-③。构造两个方程组
$$\{\begin{array}{ll}a(x_2-x_1)+b(y_2-y_1)=\frac{(x_2^2+y_2^2)-(x_1^2+y_1^2)}{2}& ①-②\\ a(x_3-x_2)+b(y_3-y_2)=\frac{(x_3^2+y_3^2)-(x_2^2+y_2^2)}{2}& ②-③\end{array}$$

我们可以给计算方程简化一下
$$\begin{array}{rl}{A}_1& =x_2-x_1\\ B_1& =y_2-y_1\\ A_2& =x_3-x_2\\ B_2& =y_3-y_2\\ C_1& =\frac{(x_2^2+y_2^2)-(x_1^2+y_1^2)}{2}\\ C_2& =\frac{(x_3^2+y_3^2)-(x_2^2+y_2^2)}{2}\end{array}$$
则可的方程组
$$\begin{gathered} A_{1}a+B_{1}b=C_1 \\ {A}_{2}a+B_{2}b=C_2 \end{gathered}$$
使用矩阵求解这个方程组
$$\left(\begin{array}{cc}{A}_1& B_1\\ A_2& B_2\end{array}\right)\left(\begin{array}{c}a\\ b\end{array}\right)=\left(\begin{array}{c}{C}_1\\ C_2\end{array}\right)$$
可得
$$\left(\begin{array}{c}a\\ b\end{array}\right)=\frac{1}{A_1{B}_2-A_2{B}_1}\left(\begin{array}{cc}{B}_2& -B_1\\ -A_2& A_1\end{array}\right)\left(\begin{array}{c}{C}_1\\ C_2\end{array}\right)$$

$$\left(\begin{array}{c}a\\ b\end{array}\right)=\left(\begin{array}{cc}\frac{B_2}{A_1{B}_2-A_2{B}_1}& \frac{-B_1}{A_1{B}_2-A_2{B}_1}\\ \frac{-A_2}{A_1{B}_2-A_2{B}_1}& \frac{A_1}{A_1{B}_2-A_2{B}_1}\end{array}\right)\left(\begin{array}{c}{C}_1\\ C_2\end{array}\right)=\left(\begin{array}{c}\frac{B_2{C}_1-B_1{C}_2}{A_1{B}_2-A_2{B}_1}\\ \frac{A_1{C}_2-A_2{C}_1}{A_1{B}_2-A_2{B}_1}\end{array}\right)$$

这样我们就得到了圆心$O(a,b)$的坐标，同时我们也发现了一个很奇特的现象
上面再做共线条判断的时候我们求了叉积
$$value=x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$$
而我们求解的a,b的分母我带入后会发现
$$A_1{B}_2-A_2{B}_1=x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$$

是不是很玄学hhhhh,有知道原理的大佬欢迎评论区解答。

代码实现

typedef struct _GL2_fPoint {float x;float y;
} GL2_fPoint;// 计算通过三点的圆的圆心和半径
bool circle_from_three_points(const GL2_fPoint& p1, const GL2_fPoint& p2, const GL2_fPoint& p3, float& cx, float& cy, float& r) {float x1 = p1.x, y1 = p1.y;float x2 = p2.x, y2 = p2.y;float x3 = p3.x, y3 = p3.y;float A = x1 * (y2 - y3) + x2 * (y3 - y1) + x3 * (y1 - y2);if (std::abs(A) < 1e-6f) {return false;}float B = (x1 * x1 + y1 * y1) * (y3 - y2) + (x2 * x2 + y2 * y2) * (y1 - y3) + (x3 * x3 + y3 * y3) * (y2 - y1);float C = (x1 * x1 + y1 * y1) * (x2 - x3) + (x2 * x2 + y2 * y2) * (x3 - x1) + (x3 * x3 + y3 * y3) * (x1 - x2);float D = (x1 * x1 + y1 * y1) * (x3 * y2 - x2 * y3) + (x2 * x2 + y2 * y2) * (x1 * y3 - x3 * y1) + (x3 * x3 + y3 * y3) * (x2 * y1 - x1 * y2);cx = -B / (2 * A);cy = -C / (2 * A);r = std::sqrt((x1 - cx) * (x1 - cx) + (y1 - cy) * (y1 - cy));return true;
}