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

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

Answer 1

[MathWorks Support Team]

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)

[Manar Mahmoud]

how i can get the boundary vertices ?

[Christoph]

Using the first approach (which is more generic anyway: really draw a circle and not a rectangle with high curvature), the boundary vertices are simply xunit and yunit.

[Walter Roberson]

Ding Li commented

useful

[bushra mughal]

can we use this code for drawing a circle on image

[Walter Roberson]

bushra mughai, see http://matlab.wikia.com/wiki/FAQ#How_do_I_create_a_circle.3F

[Chinthamakula Saikiran]

Thanks and we want answer to plot circle when they have given us centre and a point on the circle

[Michael Abboud]

If you would like to plot a circle given two points [Center, Point on circle], rather than [Center, Radius], you can simply calculate the distance between your two points, and then use that distance as the radius.

[Özgür Saglam]

Can you please explain the interval? What does pi/50 mean? Can i just write th = 0:2*pi, so the Interval goes from 0 to 2*pi

[Walter Roberson]

Yes, you can just write 0:2*pi

>> 0:2*pi

ans =

     0     1     2     3     4     5     6

This would give you 7 different radians values, and will approximate the circle as a heptagon, which might be good enough for your purposes. Each side would cover a bit less than 60 degrees. This will not at all look round to most people.

What does pi/50 mean

There are 2*pi radians in a circle, so using pi/50 as the step means that you would be approximating the circle as (2*pi)/(pi/50) = 100 sides, so each side would be roughly 3.6 degrees. That is enough sides that the circle would look curved to most people.

Answer 2

[serwan Bamerni]

There is now a function called viscircles():

http://www.mathworks.com/help/images/ref/viscircles.html?s_tid=srchtitle

[Walter Roberson]

This is part of the Image Processing Toolbox

[ceethal piyus]

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.

Answer 3

[amine bouabid]

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

Answer 4

[Supoj Choachaicharoenkul]

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

Answer 5

[Devin Marcheselli]

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

[Mark Rzewnicki]

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]

Thank you so much.

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

[Mark Rzewnicki]

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]);