Circle inside circle collision

Question

In one of my projects I have a game area in the shape of a circle. Inside this circle another small circle is moving around. What I want to do is keep the small circle from moving outside the bigger one. Below you can see that in frame 2 the small circle is partly outside, I need a way to move it back to just before it is about to move outside. How can this be done?

Also, I need the collision point along the arc of the big circle so that I can update the small circle's velocity. How would one go about calculating this point?

What I would like to do is before moving the small circle, I predict its next position and if it is outside I find the time of collision between t=0 and t=1 (t=1 full time step). If I have the collision time t then I just move the small circle during t instead of a full time step. But again, the problem is I don't know how to detect at that time the collision occurs when it comes to two circles and one being inside the other.

EDIT:

Example of collision point (green) I want to find. Maybe the picture is a bit off but you get the idea.

Answer 1

Let's assume the large circle has centre A and radius R and the small circle has centre B and radius r moving towards location C.

There is an elegant way to solve this problem, using Minkovski sums (subtractions, actually): replace the disc of radius R with a disc of radius R-r, and the disc of radius r with a disc of radius 0, ie. a simple point located at B. The problem becomes a line-circle intersection problem.

You then just need to check whether distance AC is smaller than R-r. If it is, the circles do not collide. If it is larger, just find the point D on BC at distance R-r of A and this is the new location of the centre of your small circle. This is equivalent to finding k such that:

  vec(BD) = k*vec(BC)
and
  norm(vec(AD)) = R-r

Substituting vec(AD) with vec(AB) + vec(BD) gives:

AB² + k² BC² + 2k vec(AB).vec(BC) = (R-r)²

Provided the initial position was inside the large circle, this quadratic equation in k has one positive root. Here is how to solve the equation, in pseudocode:

b = - vec(AB).vec(BC) / BC²    // dot product
c = (AB² - (R-r)²) / BC²
d = b*b - c
k = b - sqrt(d)
if (k < 0)
    k = b + sqrt(d)
if (k < 0)
    // no solution! we must be slightly out of the large circle

With this value of k, the small circle's new centre is at D such that BD = kBC.

Edit: add quadratic equation solution

Answer 2

Say the big circle is circle A and the small circle is circle B.

Check to see if B is inside A:

distance = sqrt((B.x - A.x)^2 + (B.y - A.y)^2))
if(distance > A.Radius + B.Radius) { // B is outside A }

If in frame n-1 B was inside A and in frame n B is outside A and the time between frames wasn't too big (aka B wasn't moving too fast) we can approximate the point of collision by just finding the Cartesian coordinates of B relative to A:

collision.X = B.X - A.X;
collision.Y = B.Y - A.Y;

We can then convert this points to an angle:

collision.Normalize(); //not 100% certain if this step is necessary     
radians = atan2(collision.Y, collision.X)

If you want to know more exactly at what t B is outside A for the first time you could do a ray-circle intersection every frame and then compare if the distance from B to the point of collision is bigger then the distance B can travel given it's current speed. If so you can calculate the exact time of collision.

Answer 3

Let (Xa,Ya) the position of the big circle and it's radius R, and (Xb,Yb) the position of the smaller circle and it's radius r.

You can check if these two circles collide if

DistanceTest = sqrt(((Xa - Xb) ^ 2) + ((Ya - Yb) ^ 2)) >= (R - r)

To find out the position of collision, find the exact time moment when the circles collide, by using a binary search but with a fixed number of steps. Depending on how your game is made, you can optimize this part of the code (i provided this solution to be independent of how the small ball behaves. If it has constant acceleration, or constant speed, this part of the code can be optimized and replaced with a simple formula).

left = 0 //the frame with no collision
right = 1 //the frame after collision
steps = 8 //or some other value, depending on how accurate you want it to be
while (steps > 0)
    checktime = left + (right - left) / 2
    if DistanceTest(checktime) is inside circle //if at the moment in the middle of the interval [left;right] the small circle is still inside the bigger one
        then left = checktime //the collision is after this moment of time
        else right = checktime //the collision is before
    steps -= 1
finaltime = left + (right - left) / 2 // the moment of time will be somewhere in between, so we take the moment in the middle of interval to have a small overall error

Once you know the collision time, compute the positions of the two circles at the final time and the final collision point is

CollisionX = (Xb - Xa)*R/(R-r) + Xa
CollisionY = (Yb - Ya)*R/(R-r) + Ya

Answer 4

I've implemented a demo of a ball bouncing in a circle on jsfiddle using the algorithm described by Sam Hocevar:

http://jsfiddle.net/klenwell/3ZdXf/

Here's the javascript that identifies the point of contact:

find_contact_point: function(world, ball) {
    // see https://gamedev.stackexchange.com/a/29658
    var A = world.point();
    var B = ball.point().subtract(ball.velocity());
    var C = ball.point();
    var R = world.r;
    var r = ball.r;

    var AB = B.subtract(A);
    var BC = C.subtract(B);
    var AB_len = AB.get_length();
    var BC_len = BC.get_length();

    var b = AB.dot(BC) / Math.pow(BC_len, 2) * -1;
    var c = (Math.pow(AB_len, 2) - Math.pow(R - r, 2)) / Math.pow(BC_len, 2);
    var d = b * b - c;
    var k = b - Math.sqrt(d);

    if ( k < 0 ) {
        k = b + Math.sqrt(d);
    }

    var BD = C.subtract(B);
    var BD_len = BC_len * k;
    BD.set_length(BD_len);

    var D = B.add(BD);
    return D;
}