Intersection condition between two ellipses

I have two ellipse using the ellipse equation. I need to know whether the two ellipses are intersecting or not as a Boolean value. True if intersecting, and False if not intersecting. The usage of Solve function is not working in my matlab version. So please suggest me a numerical way to write this code. I have copied the code I am using for it.

ellipseTwo = '((x+0)^2)/2^2 + ((y-5)^2)/4^2 = 1'; ellipseOne = '((x+0)^2)/2^2 + (y-5)^2/4^2 = 1';

ezplot(ellipseOne, [-10, 10, -10, 10]);

hold on

ezplot(ellipseTwo, [-10, 10, -10, 10]);

[Torsten]

Are the ellipses in standard form (i.e. their axes parallel to x- and y-axis) ?

[Sathiya]

Lets say the ellipses are in arbitrary space.

ellipseTwo = '((x+5)^2)/4^2 + ((y-2)^2)/2^2 = 1';

ellipseOne = '((x+0)^2)/4^2 + (y-5)^2/2^2 = 1';

[Sathiya]

@Torsten Yes, always the major axis and minor axis of the ellipse will be parallel to the x axis or y axis. There is no tilt in the orientation.

[Bruno Luong]

If the aspect ratio of two ellipses is the same as in the given example, you can transform the problem to two circle intersection in (x1,y1) plane with variable transformations:

% x1 = x/4
% y1 = y/2

circle intersection is easier, just compute the distance of two circle centers and compare with r1+r2.

c1 = [0; 5];  % circle center  #1
c2 = [-5; 2]; % circle center  #2
T = [1/4 0;
     0 1/2]; % Change variable matrix
C1 = T*c1;
C2 = T*c2;
r1 = 1; % radius
r2 = 1;
dC = norm(C1-C2)

dC = 1.9526

isintersecting = dC <= r1 + r2

isintersecting = logical
   1

[Sathiya]

@Bruno Luong Circle radius is not direction dependent (can be considered as a scalar line defining the radius), but in the case of ellipse, the radius(major axis or minor axis or inbetween) depends on the direction, so it is vector. Aproximating ellipse to a circle, wont be good idea. Correct me if I am wrong.

[Bruno Luong]

I don't think you get my idea. There is no approximation if tha aspect ratio of two ellipse are identical. By changing parameter, it becomes circle equation with new coordinates. This is NOT an approximation, but exact transformation.

If the aspect ratio is not identical (or radom tilted ellipses the problem become minimizing uadratic function under disc contraints; and ca, be solved with traust region algorithm. I won't go there yet since you have not confimed the asumption of aspect ratio.

Answer 1

[Torsten]

I'm surprised that we get a division by zero by the below solution method if the aspect ratio of the two ellipses is equal.

But I think sigma18 and sigma19 have a common divisor. So the problem might vanish if you use

sol = simplify(solve(e1subs,y,'MaxDegree',4))

instead of

sol = solve(e1subs,y,'MaxDegree',4)

below.

If all solutions are complex, the two ellipses don't intersect. If not, you have intersections.

If you invest a little effort, use pencil and paper and deduce the coefficients of the polynomial e1subs of degree 4. Then you can use the numerical "roots" command to deduce the possible intersections (and thus the information whether the ellipses intersect or not).

syms x y
syms x1 x2 y1 y2 real
syms a1 b1 a2 b2 real positive
e1 = (x-x1)^2/a1^2 + (y-y1)^2/b1^2 == 1;
e2 = (x-x2)^2/a2^2 + (y-y2)^2/b2^2 == 1;
solx = solve(e1*a1^2-e2*a2^2,x)

solx = 

e1subs = subs(e1,x,solx)

e1subs = 

sol = solve(e1subs,y,'MaxDegree',4)

sol = 

solsubs = subs(sol,[a1 a2 b1 b2 x1 x2 y1 y2],[4 4 2 2 0 -5 5 2])

Error using sym/subs (line 156)
Division by zero.

solsubs(1:4)

[Torsten]

There are still special cases for which the below code doesn't work, e.g. if Y1 = Y2.

% E1: (x-X1)^2/A1^2 + (y-Y1)^2/B1^2 = 1
% E2: (x-X2)^2/A2^2 + (y-Y2)^2/B2^2 = 1
%
% Numerical values for the ellipse parameters
A1 = 4;
A2 = 4;
B1 = 2;
B2 = 2;
X1 = 0;
X2 = -5;
Y1 = 5;
Y2 = 2;
%
syms x y
syms x1 x2 y1 y2 real
syms a1 b1 a2 b2 real positive
e1 = (x-x1)^2/a1^2 + (y-y1)^2/b1^2 == 1;
e2 = (x-x2)^2/a2^2 + (y-y2)^2/b2^2 == 1;
soly = solve(e1*b1^2-e2*b2^2,y);
e1subs = subs(e1,y,soly);
poly_x = fliplr(coeffs(lhs(e1subs)-rhs(e1subs),x));
sol_x = roots(subs(poly_x,[a1 a2 b1 b2 x1 x2 y1 y2],[A1 A2 B1 B2 X1 X2 Y1 Y2]));
poly_y = fliplr(coeffs(lhs(e1)-rhs(e1),y));
count = 0;
for i = 1:numel(sol_x)
  if abs(imag(sol_x(i))) < 1e-8
    sol_y = roots(subs(poly_y,[a1 a2 b1 b2 x1 x2 y1 y2 x],[A1 A2 B1 B2 X1 X2 Y1 Y2 sol_x(i)]));
    for j = 1:2
      if abs(imag(sol_y(j))) < 1e-8 & abs(subs(lhs(e2),[a1 a2 b1 b2 x1 x2 y1 y2 x y],[A1 A2 B1 B2 X1 X2 Y1 Y2 sol_x(i) sol_y(j)])-1) < 1e-8
         count = count + 1;
         SOL_X(count) = sol_x(i);
         SOL_Y(count) = sol_y(j);
      end
    end
  end
end
SOL_X

SOL_X = 

SOL_Y

SOL_Y = 

if isempty(SOL_X)
  disp('Ellipses don''t intersect')
else
  disp('Ellipses intersect')  
end

Ellipses intersect

Answer 2

[Aquatris]

Here is one way where accuracy will depend on selecting the threshold correctly. You simply solve the equation for y given an x vector, and then find the distance between the elips 1 points and elips 2 points. Then find if any distance is less than a threshold

x = -10:0.01:10;

% solve the equation for y and get index of imaginary solutions
y1 = sqrt(-((x+5).^2/4^2-1)*2^2);idx1 = imag(y1) ==0;
y2 = sqrt(-((x+0).^2/4^2-1)*2^2);idx2 = imag(y2) ==0;

% form x-y points of the elipses
x1 = x(idx1) ;x1 = [x1 flip(x1)];
y1 = y1(idx1);y1 = [y1 -flip(y1)]+2;
 
x2 = x(idx2) ;x2 = [x2 flip(x2)];
y2 = y2(idx2);y2 = [y2 -flip(y2)]+5;
 
% loopt over points and calculate their distance, this might be vectorized
for i = 1:length(x1)
   for j = 1:length(x2)
       dist(i,j) = sqrt((x1(i)-x2(j))^2+(y1(i)-y2(j))^2);
   end
end
% find indeces where distance is below a threshold
[a,b] = find(dist<2e-3);
% plot
figure(2)
plot(x1,y1,x2,y2,x1(a),y1(a),'mx',x2(b),y2(b),'ko')
axis([-10 10 -10 10])

Answer 3

[Bruno Luong]

Here is a solution for generic ellipses using FEX file here

% Ti and ci define two ellipses E1 and E2 of the form
% (x,y) st norm(Ti*Xi) <= 1 where Xi := [x;y]-ci

c1 = [0; 5];  % circle center  #1
c2 = [-5; 2]; % circle center  #2
T1 = [1/4  0;
       0  1/2]; % Change variable matrix
T2 = T1;

T = T2/T1;
A = 2*(T'*T);
dc = c1-c2;
dc2 = T2*dc;
b = 2*T*dc2;
c = dc2'*dc2;
% TrustRegion.m File from FEX
% https://www.mathworks.com/matlabcentral/fileexchange/27596-least-square-with-2-norm-constraint
Xx = TrustRegion(A, b, 1);
% X = T1\Xx + c1
f = 1/2*Xx'*A*Xx+Xx'*b+c;
isintersect = f <= 1

Answer 4

[Bruno Luong]

Use FSOLVE to find numerical solution. Only xaimum one can be found

c1 = [0; 5];  % circle center  #1
c2 = [-5; 2]; % circle center  #2
T1 = [1/4  0;
       0  1/2]; % Change variable matrix
T2 = T1;


[X, fval, exitflag] = fsolve(@(X) myfun(X,c1,T1,c2,T2), randn(2,1))

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

X = 2x1

-1.8347 3.2228

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

fval = 2x1

1.0e-07 * 0.2337 0.2337

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

exitflag = 1

isintersect = norm(fval, Inf) < 1e-6 && exitflag >= 0

isintersect = logical
   1

figure
hold on
h1=fimplicit(@(x,y) sum((T1*([x(:),y(:)]'-c1)).^2,1)-1,[-10 10]);
h2=fimplicit(@(x,y) sum((T2*([x(:),y(:)]'-c2)).^2,1)-1,[-10 10]);
h1.Color = 'k';
h2.Color = 'k';
plot(X(1),X(2),'ro','MarkerSize',10)

function F = myfun(X,c1,T1,c2,T2)
F = [sum((T1*(X-c1)).^2,1)-1;
    sum((T2*(X-c2)).^2,1)-1];
end