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.

[Tey Yi Hua]

What does r * cos(th) + x mean?

[Walter Roberson]

r * cos(th) is part of polar to cartesian coordinate conversion. For a given theta, for a circle of radius r centered on the origin, then the x coordinate is r multiplied by cos(theta) and the y coordinate is r multiplied by sin(theta). And then you add the actual center of the circle, x and y, to those coordinates.

A more common mathematical notation would be and for a circle of radius r centered on and

[Tey Yi Hua]

Okayyyy thank you very much...this help a lot!!!

[Walter Roberson]

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)

[HG]

how can I draw a triangel, rectangle and hegagon with curve edge and curve corner?

and then displaying with imagesc.

[HG]

https://www.mathworks.com/matlabcentral/answers/846710-display-geometric-shapes-and-output-matrices?s_tid=srchtitle

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 Kottakali 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.

[Walter Roberson]

viscircles(app.segmented, centres, radii, 'color', 'b')

Answer 3

[Supoj Choachaicharoenkul]

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

[wagenaartje]

This is the best solution by far if you want to highlight some part in the figure

[Walter Roberson]

Depending which graphics driver you are using, when you ask for a circle marker drawn large, the result might not look circular. The drivers approximate a circle but they do not generally take into consideration the size of the circle when doing the approximation so it might look bad.

Answer 4

[Steven Lord]

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

[HG]

how can I draw a triangel, rectangle and hegagon with curve edge and curve corner?

and then displaying with imagesc.

https://www.mathworks.com/matlabcentral/answers/846710-display-geometric-shapes-and-output-matrices?s_tid=srchtitle

[Walter Roberson]

https://matlab.fandom.com/wiki/FAQ#How_do_I_create_an_arc.3F

[Walter Roberson]

Remember that an equilateral triangle has a 60 degree range.

Answer 5

[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

[Alfonso Salas-Cruz]

Thanks, that was the shorter solution to plot circle.

[Walter Roberson]

Using viscircles() or using plot() with a 'o' marker and large 'MarkerSize' is even shorter.

Answer 6

[Ebrahim Soujeri]

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]

You can make it shorter

plotcircle(50, 40, 80)

function plotcircle(r,x,y)
plot(r*exp(j*(0:pi/100:2*pi))+x+j*y);
end

[Ebrahim Soujeri]

Sure, absolutely.

I did not want to embed though.

[Walter Roberson]

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 ;-)

Answer 7

[PATRICIA AGUILAR]

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]

See https://www.mathworks.com/matlabcentral/answers/98665-how-do-i-plot-a-circle-with-a-given-radius-and-center#answer_108013 for code to plot a circle.

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.

Answer 8

[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]);

Answer 9

[ali yaman]

I think there is no need to any of the above solutions, you can draw a circle by one of the following codes.

Let's say radius is 5

fimplicit(@(x,y) x^2+y^2-25)
ezplot('x^2+y^2-25',[-5,5])
% Note that the 25 in the codes comes from the square of 5. If you want
% draw a circle with 8 radius then write 64 instead of 25.

[Adam Danz]

One benefit of some of the other approaches is that they are faster and more efficient that this method. Also, notice the warning that appears with this method. I would be worried that users may change the values in the first line and forget to update the string in the second line of this solution.

There are often many solutions to a problem -- thanks for adding another! :)

fimplicit(@(x,y) x^2+y^2-25)

Warning: Function behaves unexpectedly on array inputs. To improve performance, properly vectorize your function to return an output with the same size and shape as the input arguments.

ezplot('x^2+y^2-25',[-5,5])

[Walter Roberson]

Caution: the default drawing area for fimplicit is [-5 5], which will not draw a circle if the radius is greater than 5.

fimplicit(@(x,y) x.^2+y.^2-6.^2, [-6.5 6.5])
axis equal

[ali yaman]

Thanks @Adam and @Walter for contribution. @Adam you do not need to use both code line to draw a circle. I said that you can draw by one of them. So there is no need to update the second code. They are diffirint, just use one of them.

[ali yaman]

Thanks @Walter Roberson. I did not know the default drawing area for fimplicit is [-5 5].