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.

1 Commento

fatima ibrahim
fatima ibrahim il 29 Feb 2020
function draw_circle1(x,y,R,c)
t =0:0.05:6.28;
x1 = (x +R*cos(t))';
y1= (x +R*sin(t))';

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 Risposta accettata

Paulo Silva
Paulo Silva il 12 Mar 2011

15 voti

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.

8 Commenti

Firas
Firas il 29 Dic 2014
That's awesome
Eddie
Eddie il 15 Mar 2017
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
Bud Kelly il 2 Apr 2018
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! ;-)
passioncoding
passioncoding il 20 Ago 2018
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
Image Analyst il 20 Ago 2018
Modificato: Image Analyst il 20 Ago 2018
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
Özgür Saglam il 3 Mag 2020
what does 0.01 in ang mean? i dont understand this part
Rik
Rik il 3 Mag 2020
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
Özgür Saglam il 3 Mag 2020
Thank you very much!

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Più risposte (3)

Michelle Hirsch
Michelle Hirsch il 29 Gen 2016

18 voti

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

5 Commenti

Image Analyst
Image Analyst il 29 Gen 2016
Modificato: Image Analyst il 15 Apr 2022
And with R2012a and later you can use viscircles
numCircles = 15;
x = 5 + randi(95, numCircles, 1);
y = 5 + randi(95, numCircles, 1);
radius = 2 * ones(numCircles, 1);
viscircles([x, y], radius);
grid on;
axis equal
Ronald Mintz
Ronald Mintz il 14 Apr 2016
Thanks very much Michelle. Your idea made beautiful concentric circles. viscircles didn't work on my computer because I have version R2011a.
Image Analyst
Image Analyst il 14 Apr 2016
Modificato: Image Analyst il 15 Apr 2022
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.
numCircles = 15;
x = 5 + randi(95, numCircles, 1);
y = 5 + randi(95, numCircles, 1);
radius = 2 * ones(numCircles, 1);
viscircles([x, y], radius);
grid on;
axis equal
Royi Avital
Royi Avital il 10 Dic 2023
@Michelle Hirsch, It would be great if it had the DisplayName property like most other objects.
It makes easier when adding it to the legend.
Michelle Hirsch
Michelle Hirsch il 11 Dic 2023
@Royi Avital I think it's more than just adding DisplayName - annotations like rectangle (intentionally) don't show up in legend since they are meant to be annotations, not data. Are you interested in being able to include annotations in legend? If so, please share more about your use case so I make sure we understand what you are thinking.

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Image Analyst
Image Analyst il 20 Gen 2016
Modificato: Image Analyst il 15 Apr 2022
numCircles = 15;
x = 5 + randi(95, numCircles, 1);
y = 5 + randi(95, numCircles, 1);
radius = 2 * ones(numCircles, 1);
viscircles([x, y], radius);
grid on;
axis equal

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