How to plot a number of circles with the same radius, but the position of the circles is randomly put

I need to plot a number of circles with the same radius. Number of circles and the radius are input parameters. Also circles are randomly positioned. For example: The program ask user to input the number of the circles, then ask to input the radius of the circle. Here, the radius is the same for all circles. Then by using input parameters program randomly put the circles.

[James Tursa]

What have you done so far? Do you know how to plot a single circle?

[Roger Stafford]

May the circles overlap? Over what range are they randomly spaced?

[Walter Roberson]

Beibit Sautbek comments,

Circles may not overlap. It should separately located

Answer 1

[KSSV]

N = 10 ;   % number of circles
r = 0.25 ; % radius of circel
%
C = rand(N,2) ;
%
th = linspace(0,2*pi) ;
x = r*cos(th) ;
y = r*sin(th) ;
% Loop for each circle
for i = 1:N
    xc = C(i,1)+x ;
    yc = C(i,2)+y ;
    hold on
    plot(xc,yc) ;
end
axis equal

[Beibit Sautbek]

Dear Dr. Siva Srinivas Kolukula, The codes is great, thank you very much. However the circles are overlap each other, I need separately located circles. Maybe I need the certain distance between the circles Thank you very much

Answer 2

[Image Analyst]

I know you already figured out how to adapt Siva's solution to prevent overlap, because you accepted it, but in case anyone else wants to know how to make a set of non-overlapping circles, see the code below:

numCircles = 1;
numCirclesMax = 50;
radius = 0.05;
maxIterations = 100 * numCirclesMax; % Fail Safe.
iteration = 1;
while  numCircles < numCirclesMax && iteration < maxIterations
  xTrial = rand;
  yTrial = rand;
  iteration = iteration + 1; % Fail safe.
  % Only need to check for overlap for second and later circles.
  if numCircles > 1
    % Find distance from other, prior circles.
    distances = sqrt((xTrial - x) .^ 2 + (yTrial - y) .^ 2);
    if min(distances) < 2 * radius
      % It's overlapping at least one of the prior ones
      continue; % Skip to end of loop and continue with loop.
    end
  end
  x(numCircles) = xTrial;
  y(numCircles) = yTrial;
  numCircles = numCircles + 1;
end
radii = radius * ones(1, length(x));
centers = [x', y'];
viscircles(centers, radii);
grid on;
axis equal

% Set up figure properties:
% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0 1 1]);
% Get rid of tool bar and pulldown menus that are along top of figure.
set(gcf, 'Toolbar', 'none', 'Menu', 'none');
% Give a name to the title bar.
set(gcf, 'Name', 'Demo by ImageAnalyst', 'NumberTitle', 'Off')

[Walter Roberson]

Beibit Sautbek comments

thank you very much.

Answer 3

[Walter Roberson]

You might be able to use viscircles() if you have the Image Processing toolbox

[jahanzaib ahmad]

using the same code i m trying to generate circle of different radius . @is it possible ?

[Image Analyst]

Well, almost the same. Of course you need to generate the random radii. See attached m-file that makes the figure below:

Answer 4

[jahanzaib ahmad]

ONE LAST THING CAN U FIX THEM IN SEMI CIRCLE .. ?

axis([-1.1 1.1 -0.1 1.1]);

[Image Analyst]

What do you mean by "fix"