Asked by MathWorks Support Team
on 26 Jul 2010

I would like to plot a circle with a given radius and center.

Answer by MathWorks Support Team
on 3 Jan 2019

Accepted Answer

Here is a MATLAB function that plots a circle with radius 'r' and locates the center at the coordinates 'x' and 'y':

function h = circle(x,y,r)

hold on

th = 0:pi/50:2*pi;

xunit = r * cos(th) + x;

yunit = r * sin(th) + y;

h = plot(xunit, yunit);

hold off

An alternative method is to use the 'rectangle' function:

function h = circle2(x,y,r)

d = r*2;

px = x-r;

py = y-r;

h = rectangle('Position',[px py d d],'Curvature',[1,1]);

daspect([1,1,1])

If you are using version R2012a or later and have Image Processing Toolbox, then you can use the 'viscircles' function to draw circles:

viscircles(centers,radii)

Answer by serwan Bamerni
on 17 Feb 2016

There is now a function called viscircles():

Walter Roberson
on 17 Oct 2016

This is part of the Image Processing Toolbox

ceethal piyus
on 16 May 2019 at 9:01

I've applied circular hough tranform for identifing the circlular objects in an image and i got the results . But when i used the code into an app designer, insted of ploting the circles into the image where the centers are marked its ploting the circle in a different figure. How can i get both circles and center points into the same axes ( named "app.segmented" )

error

How to add the axes name to viscircles function in an app designer. I tired

viscircles(centres,radii,'color','b','parent','app.segemented'); which is giving the following error.

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Answer by amine bouabid
on 23 Jul 2018

Edited by amine bouabid
on 23 Jul 2018

hello

you can plot a circle simply by writing :

syms x; syms y;

ezplot((x-xi).^2+(y-yi).^2-r.^2)

where xi and yi are the coordinates of the center and r is the radius

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## 7 Comments

## Manar Mahmoud (view profile)

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## Christoph (view profile)

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## Walter Roberson (view profile)

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## bushra mughal (view profile)

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## Walter Roberson (view profile)

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## Chinthamakula Saikiran (view profile)

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## Michael Abboud (view profile)

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