For intersecting circles, how to remove overlap and have chord visible? Also, how to randomly generate circle position in design space?

As in the image, I am trying to remove the circle overlap from the intersection, and have the chord itself (not present in the image) visible.

[Sean]

In terms of the overlap, the arc present from any circle that is inside of another circle should be removed, and the overlapped area should be removed as well.

[José-Luis]

How to ask a question

[Sara]

there's no image.

[Sean]

http://vrya.net/domain/images/img20_21.jpg

The pair of circles on the right.

[Sean]

The pair does not have the chord length (line segment connecting the two intersection points), which I have already discovered how to implement. The assistance I require is in removing the arc that overlaps another circle (the arc of one circle inside of another's boundary, or radius).

Answer 1

[Joseph Cheng]

How are you drawing your circles?

However you are trying to draw it you'll probably end up having x and y points representing the 2 circles.

xCenter1 = 7;
yCenter1 = 7;

xCenter2 = 12;
yCenter2 = 10;
theta = 0 : 0.01 : 2*pi;
radius1 = 5;
radius2 = 6;

%generate 2 circles with parameters above.
  x1 = radius1 * cos(theta) + xCenter;
  y1 = radius1 * sin(theta) + yCenter;

x2 = radius2 * cos(theta) + xCenter2;
y2 = radius2 * sin(theta) + yCenter2;
%%find distance from each circle center to the other circle's infringing points.
dC1 = sqrt((x2-xCenter1).^2+(y2-yCenter1).^2)>=radius1;
dC2 = sqrt((x1-xCenter2).^2+(y1-yCenter2).^2)>=radius2;

plot(x1(dC2), y1(dC2),'b.',x2(dC1),y2(dC1),'r.');
axis square;
xlim([0 20]);
ylim([0 20]);
grid on;

Looking at your sample picture it is easy to see that the infringing portion of circle 2 into circle 1 are points that are within the radius of circle 1. (vice versa for circle 2) With this we can calculate the distance of all points of circle 2 to the center of circle 1 and compare it to the radius.

[Joseph Cheng]

Oh and to randomly generate points you can use rand() to generate a random number.

Answer 2

[Roger Stafford]

Let P1 = [x1;y1] and P2 = [x2;y2] be column vectors for the coordinates of the two centers of the circles and let r1 and r2 be their respective radii.

d2 = sum((P2-P1).^2);
P0 = (P1+P2)/2+(r1^2-r2^2)/d2/2*(P2-P1);
t = ((r1+r2)^2-d2)*(d2-(r2-r1)^2);
if t <= 0
  fprintf('The circles don''t intersect.\n')
else
  T = sqrt(t)/d2/2*[0 -1;1 0]*(P2-P1);
  Pa = P0 + T; % Pa and Pb are circles' intersection points
  Pb = P0 - T;
  a = atan2(P1(2)-P2(2),P1(1)-P2(1));
  b1 = atan2(abs(det([Pa-P1,P1-P2])),dot(Pa-P1,P1-P2));
  b2 = atan2(abs(det([Pb-P2,P2-P1])),dot(Pb-P2,P2-P1));
  t1 = linspace(a-b1,a+b1);
  t2 = linspace(a+pi-b2,a+pi+b2);
  Q1 = bsxfun(@plus,P1,r1*[cos(t1);sin(t1)]);
  Q2 = bsxfun(@plus,P2,r2*[cos(t2);sin(t2)]);
  plot(Q1(1,:),Q1(2,:),'y-',Q2(1,:),Q2(2,:),'y-',...
       [Pa(1),Pb(1)],[Pa(2),Pb(2)],'y-')
  axis equal
end

[Sean]

Can this be enacted for more than 2 circles? If so, how?

Answer 3

[Kelly Kearney]

If you have the mapping toolbox, try polybool. If you don't, try this option, which appears to do provide a wrapper around GPC, which is the same thing polybool does.