Find intersecting point of three circles programmatically [closed]

Answer 1

You could get help from this C code. Porting it to JAVA should not be challenging. Explanation is here. Search for/scroll to: intersection of two circles

Using this method, find the intersection of any two circles.. lets say (x,y). Now the third circle will intersect at point x,y only if distance between its center and point x,y is equal to r.

case 1) If distance(center,point) == r, then x,y is the intersection point.

case 2) If distance(center,point) != r, then no such point exists.

Code (ported from [here! all credits to original author):

private boolean calculateThreeCircleIntersection(double x0, double y0, double r0,
                                                 double x1, double y1, double r1,
                                                 double x2, double y2, double r2)
{
    double a, dx, dy, d, h, rx, ry;
    double point2_x, point2_y;

    /* dx and dy are the vertical and horizontal distances between
    * the circle centers.
    */
    dx = x1 - x0;
    dy = y1 - y0;

    /* Determine the straight-line distance between the centers. */
    d = Math.sqrt((dy*dy) + (dx*dx));

    /* Check for solvability. */
    if (d > (r0 + r1))
    {
        /* no solution. circles do not intersect. */
        return false;
    }
    if (d < Math.abs(r0 - r1))
    {
        /* no solution. one circle is contained in the other */
        return false;
    }

    /* 'point 2' is the point where the line through the circle
    * intersection points crosses the line between the circle
    * centers.
    */

    /* Determine the distance from point 0 to point 2. */
    a = ((r0*r0) - (r1*r1) + (d*d)) / (2.0 * d) ;

    /* Determine the coordinates of point 2. */
    point2_x = x0 + (dx * a/d);
    point2_y = y0 + (dy * a/d);

    /* Determine the distance from point 2 to either of the
    * intersection points.
    */
    h = Math.sqrt((r0*r0) - (a*a));

    /* Now determine the offsets of the intersection points from
    * point 2.
    */
    rx = -dy * (h/d);
    ry = dx * (h/d);

    /* Determine the absolute intersection points. */
    double intersectionPoint1_x = point2_x + rx;
    double intersectionPoint2_x = point2_x - rx;
    double intersectionPoint1_y = point2_y + ry;
    double intersectionPoint2_y = point2_y - ry;

    Log.d("INTERSECTION Circle1 AND Circle2:", "(" + intersectionPoint1_x + "," + intersectionPoint1_y + ")" + " AND (" + intersectionPoint2_x + "," + intersectionPoint2_y + ")");

    /* Lets determine if circle 3 intersects at either of the above intersection points. */
    dx = intersectionPoint1_x - x2;
    dy = intersectionPoint1_y - y2;
    double d1 = Math.sqrt((dy*dy) + (dx*dx));

    dx = intersectionPoint2_x - x2;
    dy = intersectionPoint2_y - y2;
    double d2 = Math.sqrt((dy*dy) + (dx*dx));

    if(Math.abs(d1 - r2) < EPSILON) {
        Log.d("INTERSECTION Circle1 AND Circle2 AND Circle3:", "(" + intersectionPoint1_x + "," + intersectionPoint1_y + ")");
    }
    else if(Math.abs(d2 - r2) < EPSILON) {
        Log.d("INTERSECTION Circle1 AND Circle2 AND Circle3:", "(" + intersectionPoint2_x + "," + intersectionPoint2_y + ")"); //here was an error
    }
    else {
        Log.d("INTERSECTION Circle1 AND Circle2 AND Circle3:", "NONE");
    }
    return true;
}

Call this method as follows:

calculateThreeCircleIntersection(-2.0, 0.0, 2.0, // circle 1 (center_x, center_y, radius)
                                  1.0, 0.0, 1.0, // circle 2 (center_x, center_y, radius)
                                  0.0, 4.0, 4.0);// circle 3 (center_x, center_y, radius)

Also, define EPSILON to a small value that is acceptable for your application requirements

private static final double EPSILON = 0.000001;

Note: Maybe some one should test and verify if the results are correct. I can't find any easy way to do so..works for the basic cases that I have tried

Answer 2

You can use the following condition:

(x - x0) ^ 2 + (y - y0) ^ 2 <= R ^ 2

where x and y - coordinates of your point , x0 and y0 - coordinates of the center of the circle , R - radius of the circle , ^ 2 - squaring . If the condition is satisfied, the point is inside (or on the circumference in the case of equality of the left and right parts). If not satisfied, the point is outside the circle .

/ / Point, which hit the circle is necessary to determine
PointF p = ...;

/ / Center of the circle
PointF center = new PointF (10, 10);

/ / The radius of the circle
float r = 5F;

/ / "Normalize" the situation relative to the center point of the circle
float dx = p.x - center.x;
float dy = p.y - center.y;

/ / Compare the distance from the point to the center of a circle to its radius
boolean result =  ((r * r) <= (dx * dx + dy * dy))) ? true : false;