How to detect if an ellipse intersects(collides with) a circle

Question

I want to improve a collision system.

Right now I detect if 2 irregular objects collide if their bounding rectangles collide.

I want to obtain the for rectangle the corresponding ellipse while for the other one to use a circle. I found a method to obtain the ellipse coordinates but I have a problem when I try to detect if it intersects the circle.

Do you know a algorithm to test if a circle intersects an ellipse?

Answer 1

An ellipse is defined a the set of points whose sum of the distance to point A and the distance to point B is constant e. (A and B are called the foci of the ellipse).

All Points P, whose sum AP + BP is less than e, lie within the ellipse.

A circle is defined as the set of points whose distance to point C is r.

A simple test for intersection of circle and ellipse is following:

Find
P as the intersection of the circle and the line AC and
Q as the intersection of the circle and the line BC.

Circle and ellipse intersect (or the circle lies completely within the ellipse) if
AP + BP <= e   or   AQ + BQ <= e

EDIT:

After the comment of Martin DeMello and adapting my answer accordingly I thought more about the problem and found that the answer (with the 2nd check) still doesn't detect all intersections:

If circle and ellipse are intersecting only very scarcely (just a little more than being tangent) P and Q will not lie within the ellipse:

So the test described above detects collision only if the overlap is "big enough". Maybe it is good enough for your practical purposes, although mathematically it is not perfect.

Answer 2

Short answer: Solving exactly for whether the two objects intersect is complicated enough to be infeasible for the purpose of collision detection. Discretize your ellipse as an n-sided polygon for some n (depending on how accurate you need to be) and do collision detection with that polygon.

Long answer: If you insist on determining if the smooth ellipse and circle intersect, there are two main approaches. Both involve solving first for the closest point to the circle's center on the ellipse, and then comparing that distance to the circle's radius.

Approach 1: Use a parametrization of the ellipse. Transform your coordinates so that the ellipse is at the origin, with its axes aligned to the x-y axes. That is:

• Center of ellipse: (0,0)
• Center of circle: c = (cx, cy)
• Radius of circle: r
• Radius of x-aligned axis of ellipse: a
• Radius of y-aligned axis of ellipse: b.

The equation of the ellipse is then given by a cos(t), b sin(t). To find the closest point, we want to minimize the square distance || (a cos t, b sin t) - c ||^2. As Jean points out, this is "just calculus": take a derivative, and set it equal to 0. Unless I'm missing something, though, solving the resulting (quite nasty) equation for t is not possible analytically, and must be approximated using e.g. Newton's Method. Plug in the t you find into the parametric equation to get the closest point.

• Pro: Numerical solve is only in one variable, t.
• Con: You must be able to write down a parametrization of the ellipse, or transform your coordinates so that you can. This shouldn't be too hard for any reasonable representation you have of the ellipse. However, I'm going to show you a second method, which is much more general and might be useful if you have to generalize your problem to, say, 3D.

Approach 2: Use multidimensional calculus. No change of coordinates is necessary.

• Center of circle: c = (cx, cy)
• Radius of cirlce: r
• Ellipse is given by g(x, y) = 0 for a function g. For instance, per Curd's answer you might use g(x,y) = distance of (x,y) from focus 1 + distance of (x,y) from focus 2 - e.

Finding the point on the ellipse closest to the center of the circle can then be phrased as a constrained minimization problem:

Minimize ||(x,y) - c||^2 subject to g(x,y) = 0

(Minimizing the square distance is equivalent to minimizing the distance, and much more pleasant to deal with since it's a quadratic polynomial in x,y.)

To solve the constrained minimization problem, we introduce Lagrange multiplier lambda, and solve the system of equations

2 * [ (x,y) -c ] + lambda * Jg(x,y) = 0
g(x,y) = 0

Here Jg is the gradient of g. This is a system of three (nonlinear) equations in three unknowns: x, y, and lambda. We can solve this system using Newton's Method, and the (x,y) we get is the closest point to the circle's center.

• Pro: No parametrization needs to be found
• Pro: Method is very general, and works well whenever writing g is easier than finding a parametric equation (such as in 3D)
• Con: Requires a multivariable Newton solve, which is very hairy if you don't have access to a numerical method package.

Caveat: both of these approaches technically solve for the point which extremizes the distance to the circle's center. Thus the point found might be the furthest point from the circle, and not the closest. For both methods, seeding your solve with a good initial guess (the center of the circle works well for Method 2; you're on your own for Method 1) will reduce this danger.

Potential Third Approach?: It may be possible to directly solve for the roots of the system of two quadratic equations in two variables representing the circle and ellipse. If a real root exists, the objects intersect. The most direct way of solving this system, again using a numerical algorithm like Newton's Method, won't help because lack of convergence does not necessary imply nonexistence of a real root. For two quadratic equations in two variables, however, there may exist a specialized method that's guaranteed to find real roots, if they exist. I myself can't think of a way of doing this, but you may want to research it yourself (or see if someone on stackoverflow can elaborate.)

Answer 3

if a circle and an ellipse collide, then either their boundaries intersect 1, 2, 3, or 4 times(or infinitely many in the case of a circular ellipse that coincides with the circle), or the circle is within the ellipse or vice versa.

I'm assuming the circle has an equation of (x - a)^2 + (y - b)^2 <= r^2 (1) and the ellipse has an equation of [(x - c)^2]/[d^2] + [(y - e)^2]/[f^2] <= 1 (2)

To check whether one of them is inside the other, you can evaluate the equation of the circle at the coordinates of the center of the ellipse(x=c, y=e), or vice versa, and see if the inequality holds.

to check the other cases in which their boundaries intersect, you have to check whether the system of equations described by (1) and (2) has any solutions.

you can do this by adding (1) and (2), giving you

(x - a)^2 + (y - b)^2 + [(x - c)^2]/[d^2] + [(y - e)^2]/[f^2] = r^2 + 1

next you multiply out the terms, giving

x^2 - 2ax + a^2 + y^2 - 2by + b^2 + x^2/d^2 - 2cx/d^2 + c^2/d^2 + y^2/f^2 - 2ey/f^2 + e^2/f^2 = r^2 + 1

collecting like terms, we get

(1 + 1/d^2)x^2 - (2a + 2c/d^2)x + (1 + 1/f^2)y^2 - (2b + 2e/f^2)y = 1 + r^2 - a^2 - b^2 - c^2/d^2 - e^2/f^2

now let m = (1 + 1/d^2), n = -(2a + 2c/d^2), o = (1 + 1/f^2), and p = -(2b + 2e/f^2)

the equation is now mx^2 + nx + oy^2 + py = 1 + r^2 - a^2 - b^2 - c^2/d^2 - e^2/f^2

now we need to complete the squares on the left hand side

m[x^2 + (n/m)x] + o[y^2 + (p/o)y] = 1 + r^2 - a^2 - b^2 - c^2/d^2 - e^2/f^2

m[x^2 + (n/m)x + (n/2m)^2 - (n/2m)^2] + o[y^2 + (p/o)y + (p/2o)^2 - (p/2o)^2] = 1 + r^2 - a^2 - b^2 - c^2/d^2 - e^2/f^2

m[(x + n/2m)^2 - (n/2m)^2] + o[(y + p/2o)^2 - (p/2o)^2] = 1 + r^2 - a^2 - b^2 - c^2/d^2 - e^2/f^2

m(x + n/2m)^2 - m(n/2m)^2 + o(y + p/2o)^2 - o(p/2o)^2 = 1 + r^2 - a^2 - b^2 - c^2/d^2 - e^2/f^2

m(x + n/2m)^2 + o(y + p/2o)^2 = 1 + r^2 - a^2 - b^2 - c^2/d^2 - e^2/f^2 + m(n/2m)^2 + o(p/2o)^2

this system has a solution iff 1 + r^2 - a^2 - b^2 - c^2/d^2 - e^2/f^2 + m(n/2m)^2 + o(p/2o)^2 >= 0

There you have it, if I didn't make any algebraic mistakes. I don't know how much you can simplify the resulting expression, so this solution might be quite computationally expensive if you're going to check for many circles/ellipses

Answer 4

Forget about a mathematical solution. As you can easily see by by paining, you can have up to four solutions, and thus a fourth grade polynomial.

Instead just do a binary search along the edge of one of the figures. It is easy to determine if a point lies within an ellipse and even more so in a circle (just see if distance is shorter than radius).

If you really want to go for the maths, Wolfram MathWorld has a nice article here: http://mathworld.wolfram.com/Circle-EllipseIntersection.html but be warned, you'll still have to write a polynomial equation solver, probably using something like binary search.

Answer 5

Supposing: the ellipse is centred at the origin and with the semi-major axis (of length a) oriented along the x axis, and with a semi-minor axis of length b; E2 is the eccentricity squared, ie (a*a-b*b)/(a*a); the circle is centred at X,Y and of radius r.

The easy cases are: the circle centre is inside the ellipse (ie hypot(X/a, Y/b) <= 1) so there is an intersection; the circle centre is outside a circle centred at 0 of radius a+r (ie hypot(X,Y) > a+r) so there isn't an intersection.

One approach for the other cases is to compute the geodetic coordinates (latitude, height) of the circle centre. The circle intersects the ellipse if and only if the height is less than the radius.

The geodetic latitude of a point on an ellipse is the angle the normal to the ellipse at the point makes with the x axis, and the height of a point outside the ellipse is the distance of the point from the point on the ellipse closest to it. Note the geodetic latitude is not same as the polar angle from the ellipse centre to the point unless the ellipse is in fact circular.

In formulae the conversion from geodetic coordinates lat,ht to cartesian coordinates X,Y is X = (nu+ht)*cos(lat), Y = (nu * (1-E2) + ht)*sin(lat) where nu = a/sqrt( 1 - E2*sin(lat)*sin(lat)). The point on the ellipse closest to X,Y is the point with the same latitude, but zero height, ie x = nu*cos(lat), y = nu * (1-E2) * sin(lat). Note that nu is a function of latitude.

Unfortunately the process of finding lat,ht from X,Y is an iterative one. One approach is to first find the latitude, and then the height.

A little algebra shows that the latitude satisfies lat = atan2( Y+ E2*nu*sin(lat), X) which can be used to compute successive approximations to the latitude, starting at lat = atan2( Y, X*(1.0-E2)), or (more efficiently) can be solved using newton's method.

The larger E2 is, ie the flatter the ellipse is, the more iterations will be required. For example if the ellipse is nearly circular (say E2<0.1) then five iterations will get x,y below to within a*1e-12, but if the ellipse is very flat, e.g. E2=0.999 you'll need around 300 iterations to get the same accuracy!

Finally, given the latitude, the height can be computed by computing (x,y): x = nu*cos(lat), y = nu*(1-E2)*sin(lat) and then h is the distance from x,y to the circle centre, h = hypot( X-x, Y-y)