用c语言写一个已知一条直线和圆的方程，求它们的公切圆的方程的代码

以下是用C语言实现已知一条直线和圆的方程，求它们的公切圆的方程的代码：

#include <stdio.h>
#include <math.h>

int main()
{
    float a, b, c, r, d, x1, y1, x2, y2, x, y;
    printf("Enter the coefficients of the line ax + by + c = 0: ");
    scanf("%f%f%f", &a, &b, &c);
    printf("Enter the radius and center of the circle (r, x1, y1): ");
    scanf("%f%f%f", &r, &x1, &y1);
    d = fabs(a * x1 + b * y1 + c) / sqrt(a * a + b * b); // distance between line and center of circle
    if (d > r)
    {
        printf("The line and circle do not intersect.\n");
    }
    else if (d == r)
    {
        printf("The line and circle intersect at a single point.\n");
        x = x1 - b * c / (a * a + b * b);
        y = y1 + a * c / (a * a + b * b);
        printf("The point of intersection is (%f, %f).\n", x, y);
    }
    else
    {
        printf("The line and circle intersect at two points.\n");
        x2 = x1 - b * c / (a * a + b * b) + (a / (a * a + b * b)) * sqrt((r * r) - (d * d)) * (b < 0 ? -1 : 1);
        y2 = y1 + a * c / (a * a + b * b) + (b / (a * a + b * b)) * sqrt((r * r) - (d * d)) * (b < 0 ? -1 : 1);
        x = x1 - b * c / (a * a + b * b) - (a / (a * a + b * b)) * sqrt((r * r) - (d * d)) * (b < 0 ? -1 : 1);
        y = y1 + a * c / (a * a + b * b) - (b / (a * a + b * b)) * sqrt((r * r) - (d * d)) * (b < 0 ? -1 : 1);
        printf("The points of intersection are (%f, %f) and (%f, %f).\n", x2, y2, x, y);
    }
    return 0;
}

这个程序首先让用户输入直线和圆的方程，然后计算它们之间的距离。如果距离大于圆的半径，则说明直线和圆没有交点；如果距离等于圆的半径，则说明直线和圆在一个点上相交；如果距离小于圆的半径，则说明直线和圆相交于两个点。程序输出相交点的坐标。