MATLAB Answers

How do I plot a circle with a given radius and center?

11,119 views (last 30 days)
I would like to plot a circle with a given radius and center.

Accepted Answer

MathWorks Support Team
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]);
If you are using version R2012a or later and have Image Processing Toolbox, then you can use the 'viscircles' function to draw circles:
Walter Roberson
Walter Roberson on 25 Dec 2020
r*cos(theta) is used because of the definition of cos. cos(theta) = y/r where r = sqrt(x^2 + y^2) so y = r*cos(theta)

Sign in to comment.

More Answers (6)

serwan Bamerni
serwan Bamerni on 17 Feb 2016
Walter Roberson
Walter Roberson on 25 Dec 2020
viscircles(app.segmented, centres, radii, 'color', 'b')

Sign in to comment.

Supoj Choachaicharoenkul
Supoj Choachaicharoenkul on 2 Oct 2019
plot(x, y, 'bo', 'MarkerSize', 50);
  1 Comment
wagenaartje on 7 Dec 2020
This is the best solution by far if you want to highlight some part in the figure

Sign in to comment.

Steven Lord
Steven Lord on 25 Dec 2020
Another possibility is to approximate the circle using a polyshape with a large number of sides and plot that polyshape.
p = nsidedpoly(1000, 'Center', [2 3], 'Radius', 5);
plot(p, 'FaceColor', 'r')
axis equal

amine bouabid
amine bouabid on 23 Jul 2018
Edited: amine bouabid on 23 Jul 2018
you can plot a circle simply by writing :
syms x; syms y;
where xi and yi are the coordinates of the center and r is the radius

Devin Marcheselli
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
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;
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;
y = [ytemp1 ytemp2];
y1 = [ytemp11 ytemp22];
%% plot the (x,y) points
hold on
axis([-5 5 -5 5]);

Sign in to comment.

Ebrahim Soujeri
Ebrahim Soujeri on 26 Mar 2021 at 22:59
The shortest code for it could be this:
function plotcircle(r,x,y)
th = 0:pi/100:2*pi;
f = r * exp(j*th) + x+j*y;
plot(real(f), imag(f));
Walter Roberson
Walter Roberson on 27 Mar 2021 at 2:05
Notice though that I used the shortcut of plotting a single variable instead of real() and imag() of the expression. This is a "feature" of plot: if you ask to plot() a single variable and the variable is complex valued, then it uses the real component as x and the imaginary component as y. Removing the temporary variables made the code more compact, but the change to plot() only a single expression is using a different algorithm than what you used.
.. and you did say "the shortest", but my version of your approach is shorter ;-)

Sign in to comment.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!