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

MathWorks Support Team
on 3 Jan 2019

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)

serwan Bamerni
on 17 Feb 2016

There is now a function called viscircles():

ceethal piyus
on 16 May 2019

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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amine bouabid
on 23 Jul 2018

Edited: 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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Supoj Choachaicharoenkul
on 2 Oct 2019

plot(x, y, 'bo', 'MarkerSize', 50);

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Devin Marcheselli
on 17 Jan 2020

how do i plot a circle using the equation: (x-h).^2+(y-k).^2 = r.^2

Mark Rzewnicki
on 20 Jan 2020

Hey there, I've been playing around with this as it relates to complex-valued functions.

It turns out that this is a very nontrivial question. There are several issues you run into along the way - you need to make some arbitrary choices about how to define your circle. I will share with you the way I found to be most natural.

Assuming y is the indpendent variable, expand the x term in the circle equation:

Bring these expanded term to the right-hand side of the equation:

Take the square root of both sides and bring k to the right-hand side:

This is the equation you can use to plot the circle. The code I came up with is a little messy (was intended for personal use in another context) but I will throw some comments in and hopefully you find it as interesting as I did.

***The big takeaway from all of this: We can deal with circles in rectangular coordinates, but it is much more natural to use polar coordinates and think of the circle as the locus of points with distance r from the center.

% Circle equation: (x-h)^2 + (y-k)^2 = r^2

% Center: (h,k) Radius: r

h = 1;

k = 1;

r = 1;

%% In x-coordinates, the circle "starts" at h-r & "ends" at h+r

%% x_res = resolution spacing between points

xmin = h - r;

xmax = h + r;

x_res = 1e-3;

X = xmin:x_res:xmax;

%% There are 2 y-coordinates on the circle for most x-coordinates.

%% We need to duplicate every x-coordinate so we can match each x with

%% its pair of y-values.

%% Method chosen: repeat the x-coordinates as the circle "wraps around"

%% e.g.: x = [0 0.1 0.2 ... end end ... 0.2 0.1 0]

N = length(X);

x = [X flip(X)];

%% ytemp1: vector of y-values as we sweep along the circle left-to-right

%% ytemp2: vector of y-values as we sweep along the circle right-to-left

%% Whether we take positive or negative values first is arbitrary

ytemp1 = zeros(1,N);

ytemp2 = zeros(1,N);

for i = 1:1:N

square = sqrt(r^2 - X(i)^2 + 2*X(i)*h - h^2);

ytemp1(i) = k - square;

ytemp2(N+1-i) = k + square;

end

y = [ytemp1 ytemp2];

%% plot the (x,y) points

%% axis scaling coefficient c: how far past the radius should the graph go?

c = 1.5;

figure(1)

plot(x,y)

axis([h-c*r h+c*r k-c*r k+c*r]);

Ashly Tom
on 5 Mar 2020

Thank you so much.

Can you please tell how to plot 2 circles in same figure, using your code ?

Mark Rzewnicki
on 17 Mar 2020

Sadly I just saw this now, sorry.

The easiest way to do this would have been to write the original code twice (renaming the variables the second time) and plot both circles using a "hold on" statement.

This makes the code look brutally ugly - you really should vectorize things and define functions when scaling up code like this - but it will get the job done in a pinch. The result would look something like this (5-minute edit of my original code):

% Circle equation: (x-h)^2 + (y-k)^2 = r^2

% Center: (h,k) Radius: r

h = 1;

k = 1;

r = 1;

h1 = 2;

k1 = 2;

r1 = 2;

%% In x-coordinates, the circle "starts" at h-r & "ends" at h+r

%% x_res = resolution spacing between points

xmin = h - r;

xmax = h + r;

x_res = 1e-3;

X = xmin:x_res:xmax;

xmin1 = h1 - r1;

xmax1 = h1 + r1;

X1 = xmin1:x_res:xmax1;

%% There are 2 y-coordinates on the circle for most x-coordinates.

%% We need to duplicate every x-coordinate so we can match each x with

%% its pair of y-values.

%% Method chosen: repeat the x-coordinates as the circle "wraps around"

%% e.g.: x = [0 0.1 0.2 ... end end ... 0.2 0.1 0]

N = length(X);

x = [X flip(X)];

N1 = length(X1);

x1 = [X1 flip(X1)];

%% ytemp1: vector of y-values as we sweep along the circle left-to-right

%% ytemp2: vector of y-values as we sweep along the circle right-to-left

%% Whether we take positive or negative values first is arbitrary

ytemp1 = zeros(1,N);

ytemp2 = zeros(1,N);

ytemp11 = zeros(1,N1);

ytemp22 = zeros(1,N1);

for i = 1:1:N

square = sqrt(r^2 - X(i)^2 + 2*X(i)*h - h^2);

ytemp1(i) = k - square;

ytemp2(N+1-i) = k + square;

end

for i = 1:1:N1

square1 = sqrt(r1^2-X1(i)^2 + 2*X1(i)*h1 - h1^2);

ytemp11(i) = k1 - square1;

ytemp22(i) = k1 + square1;

end

y = [ytemp1 ytemp2];

y1 = [ytemp11 ytemp22];

%% plot the (x,y) points

figure(1)

plot(x,y)

hold on

plot(x1,y1)

axis([-5 5 -5 5]);

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