How to find a point in a circle excluding another lower circle?

Question

I want to get a random point in the red circle, and escaping the blue circle :

enter image description here

I made this:

double r = radius * Math.sqrt(R.nextDouble());
double theta = R.nextDouble() * 2 * Math.PI;
int x = (int) (r * Math.cos(theta));
int z = (int) (r * Math.sin(theta));

But it only find in red circle.

So, I tried to add something when X or Z are below the blue's circle radius, but I get this result :

enter image description here

I also tried to upgrade X or Z when distance between the random point and the center (which is 0/0) are too near.

I also made this snippet which test lot of X/Z position, but it will try everytimes. It's clearly not optimized because I have to filter all generated values, instead of directly generate values in required range.

function getNumberIn(min, max) {
   return Math.random() * (max - min) + min;
}

var canvas = document.getElementById("circlecanvas");
var context = canvas.getContext("2d");
context.beginPath();
context.arc(150, 150, 150, 0, Math.PI * 2, false);
context.fillStyle = "yellow";
context.fill();
context.closePath();

context.beginPath();
context.arc(150, 150, 50, 0, Math.PI * 2, false);
context.fillStyle = "red";
context.fill();
context.closePath();

context.fillStyle = "black";

let nb = 0;
for (let i = 0; i < 1000; i++) {
  let x = getNumberIn(0, 300);
  let z = getNumberIn(0, 300);
  
  let distance = Math.sqrt((x - 150) * (x - 150) + (z - 150) * (z - 150));
  if ((distance >= 50 || distance <= -50) && !(distance >= 150 || distance <= -150)) {
    context.fillRect(x, z, 3, 3);
  } else {
    nb++;
  }
}
console.log("Amount of skipped values: " + nb);

<canvas id="circlecanvas" width="300" height="300"></canvas>

How can I get random point in the red circle, excluding the blue one ?

Does this answer your question? [Generate a random point within a circle (uniformly)](https://stackoverflow.com/questions/5837572/generate-a-random-point-within-a-circle-uniformly) Just limit your range of valid radii to the range [blue_radius, red_radius] instead of [0, red_radius]. — 0x5453, Dec 14 '21 at 18:56
No @0x5453 , I mention the code from this and I wrote this code in my question — Elikill58, Dec 14 '21 at 19:03
Don't you work with random variables ? https://en.wikipedia.org/wiki/Probability_distribution — Yves Daoust, Dec 14 '21 at 19:22
@YvesDaoust Yes I work with random variables, I want a random position in the circle — Elikill58, Dec 14 '21 at 19:24
@YvesDaoust If I well understand, it's a uniform distribution. It's the `Random.nextInt` java's function — Elikill58, Dec 14 '21 at 20:09
This is the distribution of the generator uniform in an interval. Not the distribution of your 2D points. Do you confirm uniform in a ring ? — Yves Daoust, Dec 14 '21 at 20:30

MBo · Accepted Answer · 2021-12-14T19:24:20.543

2

Do the next change:

double r = Math.sqrt(
                     r_small*r_small + 
                     R.nextDouble() * (r_large*r_large-r_small*r_small));
//next lines remain the same 
double theta = R.nextDouble() * 2 * Math.PI;
int x = (int) (r * Math.cos(theta));
int z = (int) (r * Math.sin(theta));

where r_small is blue radius and r_large is outer radius of the red ring

This approach provides proper minimal and maximal distances and uniform distribution of points in the red ring.

edited Dec 14 '21 at 19:24

answered Dec 14 '21 at 19:16

MBo

72,080
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And so, I have to calculate this for the X then for the Z ? – Elikill58 Dec 14 '21 at 19:21
This is for radius, three your lines for theta, x and z are right. – MBo Dec 14 '21 at 19:23

How to find a point in a circle excluding another lower circle?

1 Answers1