Points on circle separated by given distance

Question

First of all, I am not looking to for points spaced uniformly around a circle, I know that has been answered many times. Instead, I have one point on a circle, and I need to find another that is a certain distance from it.

Here is an illustration : enter image description here

The distance can be either between the two points (black dotted line), or the length of the circumference between the points (blue line), whatever is simplest (accuracy is not very important).

I know the following variables:

(green point x, y)
d
r
(centre point x, y)

So how can I find one of the red points?

Well, I am using this math to do calculations for a 3D game. Is there somewhere that would be better to ask? — Ben Cracknell, Aug 03 '13 at 20:22
I've checked my function against several examples on the net, and it seems to give correct answers. I just hope it won't be too hard to turn into c#. — MightyPork, Aug 03 '13 at 21:46
Btw, if you could add tag "java", my answer would get syntax highlighting. — MightyPork, Aug 03 '13 at 22:01

score 3 · Accepted Answer · edited May 23 '17 at 12:19

So, basically you want to get intersection points of two circles:

The big one (BluePoint, radius = R)
A small one (GreenPoint, radius = D)

enter image description here
(Please excuse my amazing drawing skills :P)

I've at first tried to solve it myself, and fruitlessly wasted several sheets of paper.

Then I started googling and found an algorithm in other question.

Here is my Java implementation

double[][] getCircleIntersection(
        double x0, double y0, double r0,
        double x1, double y1, double r1) {

    // dist of centers
    double d = sqrt(sq(x0 - x1) + sq(y0 - y1));

    if (d > r0 + r1) return null; // no intersection
    if (d < abs(r0 - r1)) return null; // contained inside

    double a = (sq(r0) - sq(r1) + sq(d)) / (2 * d);

    double h = sqrt(sq(r0) - sq(a));

    // point P2
    double x2 = x0 + a * (x1 - x0) / d;
    double y2 = y0 + a * (y1 - y0) / d;

    // solution A
    double x3_A = x2 + h * (y1 - y0) / d;
    double y3_A = y2 - h * (x1 - x0) / d;

    // solution B
    double x3_B = x2 - h * (y1 - y0) / d;
    double y3_B = y2 + h * (x1 - x0) / d;

    return new double[][] {
        { x3_A, y3_A },
        { x3_B, y3_B }
    };
}

// helper functions
double sq(double val) {
    return Math.pow(val, 2);
}

double sqrt(double val) {
    return Math.sqrt(val);
}

double abs(double val) {
    return Math.abs(val);
}

This is how you would use it for the question situation:

double centerX = 0;
double centerY = 0;
double radius = 5;

double pointX = 10;
double pointY = 0;
double newPointDist = 5;        

double[][] points = getCircleIntersection(centerX, centerY, radius, pointX, pointY, newPointDist);

System.out.println("A = [" + points[0][0] + " , " + points[0][3] + "]");
System.out.println("B = [" + points[1][0] + " , " + points[1][4] + "]");

That looks great. I'll be converting it to c# and testing it soon. — Ben Cracknell, Aug 03 '13 at 22:06

score 0 · Answer 2 · answered Aug 03 '13 at 21:38

0

enter image description here

Project your right red point down on both axes to get X and Y.

From there, you'll get 2 distinct right angle triangles:

equation setup

Solutions:

equation solution

answered Aug 03 '13 at 21:38

Daniel Mošmondor

19,718
12
58
99

1

Am I wrong, or will this work **only** if the situation is axis-aligned? – MightyPork Aug 03 '13 at 21:43
1

I like your solution, it looks pretty and easy - but I'm afraid it won't work for this case: Blue and Green points can apparently be anywhere in the 2D space, and rotating coordinate system is really painful. – MightyPork Aug 03 '13 at 21:49

Points on circle separated by given distance

2 Answers2