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)

[Walter Roberson]

Ding Li commented

useful

[Walter Roberson]

bushra mughai, see https://matlab.fandom.com/wiki/FAQ#How_do_I_create_a_circle.3F

[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.

[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.

[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

[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)

[Image Analyst]

@Yuvin Wickramanayake no. It means that you put that code into a script, maybe it was called testcircle.m and preceding that code you had code to call the function like

h = circle(x, y, r);

At least that's one possibility. If so, you'd need to close your circle() function with the "end" keyword, which will allow a function to be defined and called from within a script.

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
end

If you still have trouble, attach your whole m-file in a NEW question (not here).

[Walter Roberson]

You would get a different error message if you had a script with a function after and the function did not have a matching end statement.

That error could happen if you try to create a function inside a script in a MATLAB version before R2015b. It might perhaps also happen if you try to define a function at the command line (I seem to remember the wording as being slightly different for that case, but perhaps that is the wording in an older version than I am using.)

Answer 2

[serwan Bamerni]

There is now a function called viscircles():

https://www.mathworks.com/help/images/ref/viscircles.html

[Walter Roberson]

This is part of the Image Processing Toolbox

[Walter Roberson]

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

Answer 3

[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

[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 4

[Shengchao Lin]

Using function "fplot" would be my simplest way:

Unit circle:

fplot(@(t) sin(t), @(t) cos(t));
grid on

Circle with center "x", "y" and radius "r":

x = 3; y = 4; r = 5;
fplot(@(t) r*sin(t)+x, @(t) r*cos(t)+y);
grid on;