Task: Circles of given radius through two points
C: #include<stdio.h>
#include<math.h>

typedef struct{
	double x,y;
	}point;
	
double distance(point p1,point p2)
{
	return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
	
void findCircles(point p1,point p2,double radius)
{
	double separation = distance(p1,p2),mirrorDistance;
	
	if(separation == 0.0)
	{
		radius == 0.0 ? printf("\nNo circles can be drawn through (%.4f,%.4f)",p1.x,p1.y):
							 printf("\nInfinitely many circles can be drawn through (%.4f,%.4f)",p1.x,p1.y);
	}
	
	else if(separation == 2*radius)
	{
		printf("\nGiven points are opposite ends of a diameter of the circle with center (%.4f,%.4f) and radius %.4f",(p1.x+p2.x)/2,(p1.y+p2.y)/2,radius); 
	}
	
	else if(separation > 2*radius)
	{
		printf("\nGiven points are farther away from each other than a diameter of a circle with radius %.4f",radius);
	}   
	
	else
	{
		mirrorDistance =sqrt(pow(radius,2) - pow(separation/2,2));
		
		printf("\nTwo circles are possible.");
		printf("\nCircle C1 with center (%.4f,%.4f), radius %.4f and Circle C2 with center (%.4f,%.4f), radius %.4f",(p1.x+p2.x)/2 + mirrorDistance*(p1.y-p2.y)/separation,(p1.y+p2.y)/2 + mirrorDistance*(p2.x-p1.x)/separation,radius,(p1.x+p2.x)/2 - mirrorDistance*(p1.y-p2.y)/separation,(p1.y+p2.y)/2 - mirrorDistance*(p2.x-p1.x)/separation,radius);
	}
}

int main()
{
    int i;

    point cases[] = 	
    {	{0.1234, 0.9876},    {0.8765, 0.2345},  
	{0.0000, 2.0000},    {0.0000, 0.0000},   
	{0.1234, 0.9876},    {0.1234, 0.9876},   
	{0.1234, 0.9876},    {0.8765, 0.2345},    
	{0.1234, 0.9876},    {0.1234, 0.9876}
    };

    double radii[] = {2.0,1.0,2.0,0.5,0.0};

    for(i=0;i<5;i++)
    {	
	printf("\nCase %d)",i+1);
	findCircles(cases[2*i],cases[2*i+1],radii[i]);
    }

    return 0;
}

C++: #include <iostream>
#include <cmath>
#include <tuple>

struct point { double x, y; };

bool operator==(const point& lhs, const point& rhs)
{ return std::tie(lhs.x, lhs.y) == std::tie(rhs.x, rhs.y); }

enum result_category { NONE, ONE_COINCEDENT, ONE_DIAMETER, TWO, INFINITE };

using result_t = std::tuple<result_category, point, point>;

double distance(point l, point r)
{ return std::hypot(l.x - r.x, l.y - r.y); }

result_t find_circles(point p1, point p2, double r)
{
    point ans1 { 1/0., 1/0.}, ans2 { 1/0., 1/0.};
    if (p1 == p2) {
        if(r == 0.) return std::make_tuple(ONE_COINCEDENT, p1,   p2  );
        else        return std::make_tuple(INFINITE,       ans1, ans2);
    }
    point center { p1.x/2 + p2.x/2, p1.y/2 + p2.y/2};
    double half_distance = distance(center, p1);
    if(half_distance > r)      return std::make_tuple(NONE,         ans1,   ans2);
    if(half_distance - r == 0) return std::make_tuple(ONE_DIAMETER, center, ans2);
    double root = std::hypot(r, half_distance) / distance(p1, p2);
    ans1.x = center.x + root * (p1.y - p2.y);
    ans1.y = center.y + root * (p2.x - p1.x);
    ans2.x = center.x - root * (p1.y - p2.y);
    ans2.y = center.y - root * (p2.x - p1.x);
    return std::make_tuple(TWO, ans1, ans2);
}

void print(result_t result, std::ostream& out = std::cout)
{
    point r1, r2; result_category res;
    std::tie(res, r1, r2) = result;
    switch(res) {
      case NONE:
        out << "There are no solutions, points are too far away\n"; break;
      case ONE_COINCEDENT: case ONE_DIAMETER:
        out << "Only one solution: " << r1.x << ' ' << r1.y << '\n'; break;
      case INFINITE:
        out << "Infinitely many circles can be drawn\n"; break;
      case TWO:
        out << "Two solutions: " << r1.x << ' ' << r1.y << " and " << r2.x << ' ' << r2.y << '\n'; break;
    }
}

int main()
{
    constexpr int size = 5;
    const point points[size*2] = {
        {0.1234, 0.9876}, {0.8765, 0.2345}, {0.0000, 2.0000}, {0.0000, 0.0000},
        {0.1234, 0.9876}, {0.1234, 0.9876}, {0.1234, 0.9876}, {0.8765, 0.2345},
        {0.1234, 0.9876}, {0.1234, 0.9876}
    };
    const double radius[size] = {2., 1., 2., .5, 0.};

    for(int i = 0; i < size; ++i)
        print(find_circles(points[i*2], points[i*2 + 1], radius[i]));
}