不共线的三个点可以确定一个圆。

下图是用Graph画的一个很粗糙的图片用于方便理解。 紫色的线分别为AB的中垂线和AC的中垂线。 两个中垂线的交点就是圆心。 圆心到三个点的距离都相等。

推导公式

参考博客：三点确定一个圆的计算方法

设圆心坐标O为

(

x

0

,

y

0

)

(x_0, y_0)

(x0​,y0​)，半径为r 三个点的坐标分别是，A

(

x

1

,

y

1

)

(x_1, y_1)

(x1​,y1​)，B

(

x

2

,

y

2

)

(x_2, y_2)

(x2​,y2​)，C

(

x

3

,

y

3

)

(x_3, y_3)

(x3​,y3​)

三个点到圆心的距离相等

{

(

x

1

−

x

0

)

2

+

(

y

1

−

y

0

)

2

=

r

2

(

x

2

−

x

0

)

2

+

(

y

2

−

y

0

)

2

=

r

2

(

x

3

−

x

0

)

2

+

(

y

3

−

y

0

)

2

=

r

2

\left\{ \begin{array}{c} (x_1−x_0)^2+(y_1−y_0)^2=r^2 \\ \\ (x_2−x_0)^2+(y_2−y_0)^2=r^2 \\ \\ (x_3−x_0)^2+(y_3−y_0)^2=r^2 \end{array} \right.

⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧​(x1​−x0​)2+(y1​−y0​)2=r2(x2​−x0​)2+(y2​−y0​)2=r2(x3​−x0​)2+(y3​−y0​)2=r2​ 化简得到：

(

x

1

−

x

2

)

x

0

+

(

y

1

−

y

2

)

y

0

=

(

x

1

2

−

x

2

2

)

−

(

y

2

2

−

y

1

2

)

2

(x_1-x_2)x_0 + (y_1-y_2)y_0 = \frac{(x_1^2-x_2^2)-(y_2^2-y_1^2)}{2}

(x1​−x2​)x0​+(y1​−y2​)y0​=2(x12​−x22​)−(y22​−y12​)​

(

x

1

−

x

3

)

x

0

+

(

y

1

−

y

3

)

y

0

=

(

x

1

2

−

x

3

2

)

−

(

y

3

2

−

y

1

2

)

2

(x_1-x_3)x_0 + (y_1-y_3)y_0 = \frac{(x_1^2-x_3^2)-(y_3^2-y_1^2)}{2}

(x1​−x3​)x0​+(y1​−y3​)y0​=2(x12​−x32​)−(y32​−y12​)​

使用克拉默法则对行列式求解

∣

A

∣

=

∣

(

x

1

−

x

2

)

(

y

1

−

y

2

)

(

x

1

−

x

3

)

(

y

1

−

y

3

)

∣

\begin{vmatrix} A \end{vmatrix} = \begin{vmatrix} (x_1-x_2) & (y_1-y_2) \\ \\ (x_1-x_3) & (y_1-y_3) \\ \end{vmatrix}

∣∣​A​∣∣​=∣∣∣∣∣∣​(x1​−x2​)(x1​−x3​)​(y1​−y2​)(y1​−y3​)​∣∣∣∣∣∣​

∣

b

∣

=

∣

(

x

1

2

−

x

2

2

)

−

(

y

2

2

−

y

1

2

)

2

(

x

1

2

−

x

3

2

)

−

(

y

3

2

−

y

1

2

)

2

∣

\begin{vmatrix} b \end{vmatrix} = \begin{vmatrix} \cfrac{(x_1^2-x_2^2)-(y_2^2-y_1^2)}{2} \\ \\ \cfrac{(x_1^2-x_3^2)-(y_3^2-y_1^2)}{2} \\ \end{vmatrix}

∣∣​b​∣∣​=∣∣∣∣∣∣∣∣∣​2(x12​−x22​)−(y22​−y12​)​2(x12​−x32​)−(y32​−y12​)​​∣∣∣∣∣∣∣∣∣​

∣

A

1

∣

=

∣

(

x

1

2

−

x

2

2

)

−

(

y

2

2

−

y

1

2

)

2

(

y

1

−

y

2

)

(

x

1

2

−

x

3

2

)

−

(

y

3

2

−

y

1

2

)

2

(

y

1

−

y

3

)

∣

\begin{vmatrix} A_1 \end{vmatrix} = \begin{vmatrix} \cfrac{(x_1^2-x_2^2)-(y_2^2-y_1^2)}{2} & (y_1-y_2) \\ \\ \cfrac{(x_1^2-x_3^2)-(y_3^2-y_1^2)}{2} & (y_1-y_3) \\ \end{vmatrix}

∣∣​A1​​∣∣​=∣∣∣∣∣∣∣∣∣​2(x12​−x22​)−(y22​−y12​)​2(x12​−x32​)−(y32​−y12​)​​(y1​−y2​)(y1​−y3​)​∣∣∣∣∣∣∣∣∣​

∣

A

2

∣

=

∣

(

x

1

−

x

2

)

(

x

1

2

−

x

2

2

)

−

(

y

2

2

−

y

1

2

)

2

(

x

1

−

x

3

)

(

x

1

2

−

x

3

2

)

−

(

y

3

2

−

y

1

2

)

2

∣

\begin{vmatrix} A_2 \end{vmatrix} = \begin{vmatrix} (x_1-x_2) & \cfrac{(x_1^2-x_2^2)-(y_2^2-y_1^2)}{2} \\ \\ (x_1-x_3) & \cfrac{(x_1^2-x_3^2)-(y_3^2-y_1^2)}{2} \\ \end{vmatrix}

∣∣​A2​​∣∣​=∣∣∣∣∣∣∣∣∣​(x1​−x2​)(x1​−x3​)​2(x12​−x22​)−(y22​−y12​)​2(x12​−x32​)−(y32​−y12​)​​∣∣∣∣∣∣∣∣∣​

x

1

=

∣

A

1

∣

∣

A

∣

x_1 = \cfrac{\begin{vmatrix}A_1\end{vmatrix}}{\begin{vmatrix}A\end{vmatrix}}

x1​=∣∣​A​∣∣​∣∣​A1​​∣∣​​ ,

x

2

=

∣

A

2

∣

∣

A

∣

x_2 = \cfrac{\begin{vmatrix}A_2\end{vmatrix}}{\begin{vmatrix}A\end{vmatrix}}

x2​=∣∣​A​∣∣​∣∣​A2​​∣∣​​

令

a

=

x

1

−

x

2

b

=

y

1

−

y

2

;

c

=

x

1

−

x

3

;

d

=

y

1

−

y

3

;

e

=

(

x

1

2

−

x

2

2

)

−

(

y

2

2

−

y

1

2

)

2

f

=

(

x

1

2

−

x

3

2

)

−

(

y

3

2

−

y

1

2

)

2

a = x_1 - x_2 \\ b = y_1 - y_2; \\ c = x_1 - x_3; \\ d = y_1 - y_3; \\ e = \cfrac{(x_1^2-x_2^2)-(y_2^2-y_1^2)}{2} \\ f = \cfrac{(x_1^2-x_3^2)-(y_3^2-y_1^2)}{2}

a=x1​−x2​b=y1​−y2​;c=x1​−x3​;d=y1​−y3​;e=2(x12​−x22​)−(y22​−y12​)​f=2(x12​−x32​)−(y32​−y12​)​

则

x

=

e

d

−

b

f

a

d

−

b

c

x = \cfrac{ed-bf}{ad-bc}

x=ad−bced−bf​

y

=

a

f

−

e

c

a

d

−

b

c

y = \cfrac{af-ec}{ad-bc}

y=ad−bcaf−ec​

void function(double x1, double y1, double x2, double y2, double x3, double y3){

double a = x1 - x2;

double b = y1 - y2;

double c = x1 - x3;

double d = y1 - y3;

double e = ((x1*x1-x2*x2)-(y2*y2-y1*y1))/2;

double f = ((x1*x1-x3*x3)-(y3*y3-y1*y1))/2;

// 圆心位置

x = (e*d - b*f)/(a*d - b*c);

y = (a*f - e*c)/(a*d - b*c);

}