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 Kommentar

fatima ibrahim
fatima ibrahim am 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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 Akzeptierte Antwort

Paulo Silva
Paulo Silva am 12 Mär. 2011

15 Stimmen

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 Kommentare

Firas
Firas am 29 Dez. 2014
That's awesome
Eddie
Eddie am 15 Mär. 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 am 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 am 20 Aug. 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 am 20 Aug. 2018
Bearbeitet: Image Analyst am 20 Aug. 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 am 3 Mai 2020
what does 0.01 in ang mean? i dont understand this part
Rik
Rik am 3 Mai 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 am 3 Mai 2020
Thank you very much!

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Weitere Antworten (3)

Michelle Hirsch
Michelle Hirsch am 29 Jan. 2016

18 Stimmen

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 Kommentare

Image Analyst
Image Analyst am 29 Jan. 2016
Bearbeitet: Image Analyst am 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 am 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 am 14 Apr. 2016
Bearbeitet: Image Analyst am 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 am 10 Dez. 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 am 11 Dez. 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 am 20 Jan. 2016
Bearbeitet: Image Analyst am 15 Apr. 2022

0 Stimmen

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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