Plot x^2+y^2=4

Hello, I have a little starter question about matlab. How do I plot a circle given by x^2+y^2=4?

Thank you.

Answer 1

[Sky Sartorius]

There are a few ways to go about this. One that is somewhat agnostic to what the equation is trying to represent (in this case, a circle) involves calculating the equation for the whole space, then plotting only an isoline of the target value.

[X,Y] = meshgrid(-3:.1:3,-3:.1:3); % Generate domain.
Z = X.^2 + Y.^2; % Find function value everywhere in the domain.
contour(X,Y,Z,[4 4]) % Plot the isoline where the function value is 4.

If you know more about your function and can turn it around into a function of only one variable (e.g., sine and cosine of t), that is preferable in most cases.

[Mohamed Lawindy]

Thank you very much.

Answer 2

[James Tursa]

E.g., since you know it is a circle with radius 2 centered at the origin;

ang = 0:0.01:2*pi;
x = 2*cos(ang);
y = 2*sin(ang);
plot(x,y);

[Mohamed Lawindy]

Thank you too.

Answer 3

[hamza]

Plot the contour plots of the circles x^2+y^2 of radius 1,2, 1.41,1.73.

[Image Analyst]

radii = [1, 2, 1.41, 1.73];
viscircles([zeros(4,1), zeros(4,1)], radii);
axis equal
grid on;

Answer 4

[Steven Lord]

Another way to do this is to use the fcontour function.

f = @(x, y) x.^2+y.^2;
fcontour(f, 'LevelList', 4)
axis equal

If you want to see multiple contours, specify a non-scalar LevelList.

figure
fcontour(f, 'LevelList', 1:4:25)
axis equal

[Image Analyst]

And yet another way

viscircles([0,0], 2)

ans = 
  Group with properties:

    Children: [2×1 Line]
     Visible: on
     HitTest: on

  Show all properties

[Steven Lord]

Note that viscircles is part of Image Processing Toolbox which means that not all users would have access to it.