plotting circles

How can I plot circles, same radius and different centers, all in one graph. I used the following command to draw +,o,diamond: plot (x,y,'ro',u,v,'gd',A,B,'b+'); where x,y,u,v,A,B are all row vectors. And I want to add circles to that plot where the o will be the center.

[fatima ibrahim]

function draw_circle1(x,y,R,c)

t =0:0.05:6.28;

x1 = (x +R*cos(t))';

y1= (x +R*sin(t))';

Answer 1

[Paulo Silva]

Here's a function to draw circles:

function circle(x,y,r)
%x and y are the coordinates of the center of the circle
%r is the radius of the circle
%0.01 is the angle step, bigger values will draw the circle faster but
%you might notice imperfections (not very smooth)
ang=0:0.01:2*pi; 
xp=r*cos(ang);
yp=r*sin(ang);
plot(x+xp,y+yp);
end

If you want to add circles you must insert the command

hold on

before the circles being added.

[Firas]

That's awesome

[Eddie]

Thanks a lot, this is great. If I had some time, I would probably tweak your code and try to plot open intervals on the real line, that's what I really need, but circles are fine too.

[Bud Kelly]

There are coding solutions and there are coding solutions. The ones that accomplish the task as this one does are...elegant... beautiful... art. Thank you, you may consider it stolen! ;-)

[faiza gul]

I have been trying to create function for circle with function that mentioned above ,but I am getting this "Not enough input arguments" for "xp=r*cos(ang)". kindly help me in this regard

[Image Analyst]

What is ang for you? Did you comment out or delete the line

ang=0:0.01:2*pi;

??? You did one or the other because if you didn't, ang would exist and you would not get that error.

Do you know how to debug by setting a breakpoint on that line and looking at its value? If not, see this link

[Özgür Saglam]

what does 0.01 in ang mean? i dont understand this part

[Rik]

It's the step size. You can use the colon in two ways when you create an array:

start:stop
start:step:stop

[Özgür Saglam]

Thank you very much!

Answer 2

[Michelle Hirsch]

It's counter-intuitive, but this is actually really easy with the rectangle function. From the rectangle documentation :

   pos = [2 4 2 2];
   rectangle('Position',pos,'Curvature',[1 1])
   axis equal

[Image Analyst]

And with R2012a and later you can use viscircles

[Ronald Mintz]

Thanks very much Michelle. Your idea made beautiful concentric circles. viscircles didn't work on my computer because I have version R2011a.

[Image Analyst]

rectangle() is one of several methods listed in the FAQ: http://matlab.wikia.com/wiki/FAQ#How_do_I_create_a_circle.3F> You'll have lots of other nice improvements that they've made over the last 5 years if you upgrade.

Answer 3

[Image Analyst]

Please refer to the FAQ: http://matlab.wikia.com/wiki/FAQ#How_do_I_create_a_circle.3F

Answer 4

[Chad Greene]

This'll do the job.

Answer 5

[Image Analyst]

There is now a function called viscircles(): http://www.mathworks.com/help/images/ref/viscircles.html?s_tid=srchtitle