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;
}

VB: Public Sub circles()
    tests = [{0.1234, 0.9876, 0.8765, 0.2345, 2.0; 0.0000, 2.0000, 0.0000, 0.0000, 1.0; 0.1234, 0.9876, 0.1234, 0.9876, 2.0; 0.1234, 0.9876, 0.8765, 0.2345, 0.5; 0.1234, 0.9876, 0.1234, 0.9876, 0.0}]
    For i = 1 To UBound(tests)
        x1 = tests(i, 1)
        y1 = tests(i, 2)
        x2 = tests(i, 3)
        y2 = tests(i, 4)
        R = tests(i, 5)
        xd = x2 - x1
        yd = y1 - y2
        s2 = xd * xd + yd * yd
        sep = Sqr(s2)
        xh = (x1 + x2) / 2
        yh = (y1 + y2) / 2
        Dim txt As String
        If sep = 0 Then
            txt = "same points/" & IIf(R = 0, "radius is zero", "infinite solutions")
        Else
            If sep = 2 * R Then
                txt = "opposite ends of diameter with centre " & xh & ", " & yh & "."
            Else
                If sep > 2 * R Then
                    txt = "too far apart " & sep & " > " & 2 * R
                Else
                    md = Sqr(R * R - s2 / 4)
                    xs = md * xd / sep
                    ys = md * yd / sep
                    txt = "{" & Format(xh + ys, "0.0000") & ", " & Format(yh + xs, "0.0000") & _
                    "} and {" & Format(xh - ys, "0.0000") & ", " & Format(yh - xs, "0.0000") & "}"
                End If
            End If
        End If
        Debug.Print "points " & "{" & x1 & ", " & y1 & "}" & ", " & "{" & x2 & ", " & y2 & "}" & " with radius " & R & " ==> " & txt
    Next i
End Sub