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

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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:
HG on 9 Jun 2021

serwan Bamerni on 17 Feb 2016
Walter Roberson on 25 Dec 2020

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

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
Walter Roberson on 9 Jun 2021
Remember that an equilateral triangle has a 60 degree range.

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
##### 2 CommentsShowHide 1 older comment
Walter Roberson on 9 May 2021
Using viscircles() or using plot() with a 'o' marker and large 'MarkerSize' is even shorter.

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 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
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]);

Ebrahim Soujeri on 26 Mar 2021
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 on 27 Mar 2021
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 ;-)

PATRICIA AGUILAR on 4 May 2021
An object moves on a circle of radius 1. Plot this circle and place a point at an angle of 67º. Help me
Walter Roberson on 4 May 2021
For 67 degrees, notice that in
xunit = r * cos(th) + x;
yunit = r * sin(th) + y;
you could use cosd() and sind() if your angle were in degrees.