求任意两个圆的交点，c++

方法

联立方程组，使用matlab的solve求解符号方程组;
化简表达式；
编写C++代码。

方程组

$$\{\begin{array}{l}(x-a_1{)}^2+(y-b_1{)}^2=R_1^2\\ (x-a_2{)}^2+(y-b_2{)}^2=R_2^2\end{array}$$

matlab求解代码：

clear all
syms a1 b1 R1 a2 b2 R2 x y;
[x,y]=solve((x-a1)*(x-a1)+(y-b1)*(y-b1)-R1*R1,(x-a2)*(x-a2)+(y-b2)*(y-b2)-R2*R2);
%simplify(x)
%simplify(y)

再用simplify()化简表达式，得到解析解。

C++实现

typedef struct
{
	float x;
	float y;
}Point2f;//类似opencv

typedef struct
{
	Point2f c;
	float r;
}CIRCLE;

void IntersectionOf2Circles(CIRCLE c1, CIRCLE c2, Point2f &P1, Point2f &P2)
{
	float a1, b1, R1, a2, b2, R2;
	a1 = c1.c.x;
	b1 = c1.c.y;
	R1 = c1.r;

	a2 = c2.c.x;
	b2 = c2.c.y;
	R2 = c2.r;

	//
	float R1R1 = R1*R1;
	float a1a1 = a1*a1;
	float b1b1 = b1*b1;

	float a2a2 = a2*a2;
	float b2b2 = b2*b2;
	float R2R2 = R2*R2;

	float subs1 = a1a1 - 2 * a1*a2 + a2a2 + b1b1 - 2 * b1*b2 + b2b2;
	float subs2 = -R1R1 * a1 + R1R1 * a2 + R2R2 * a1 - R2R2 * a2 + a1a1*a1 - a1a1 * a2 - a1*a2a2 + a1*b1b1 - 2 * a1*b1*b2 + a1*b2b2 + a2a2*a2 + a2*b1b1 - 2 * a2*b1*b2 + a2*b2b2;
	float subs3 = -R1R1 * b1 + R1R1 * b2 + R2R2 * b1 - R2R2 * b2 + a1a1*b1 + a1a1 * b2 - 2 * a1*a2*b1 - 2 * a1*a2*b2 + a2a2 * b1 + a2a2 * b2 + b1b1*b1 - b1b1 * b2 - b1*b2b2 + b2b2*b2;
	float sigma = sqrt((R1R1 + 2 * R1*R2 + R2R2 - a1a1 + 2 * a1*a2 - a2a2 - b1b1 + 2 * b1*b2 - b2b2)*(-R1R1 + 2 * R1*R2 - R2R2 + subs1));
	if(abs(subs1)>0.0000001)//分母不为0
	{
		P1.x = (subs2 - sigma*b1 + sigma*b2) / (2 * subs1);
		P2.x = (subs2 + sigma*b1 - sigma*b2) / (2 * subs1);
	
		P1.y = (subs3 + sigma*a1 - sigma*a2) / (2 * subs1);
		P2.y = (subs3 - sigma*a1 + sigma*a2) / (2 * subs1);
	}

}

完成！第一次发csdn博客！！