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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)

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

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

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

% 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]);

Ebrahim Soujeri
about 3 hours ago

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
3 minutes ago

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 ;-)

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