Plot x^2+y^2=4

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Mohamed Lawindy
Mohamed Lawindy 2020-2-25
Hello, I have a little starter question about matlab. How do I plot a circle given by x^2+y^2=4?
Thank you.

采纳的回答

Sky Sartorius
Sky Sartorius 2020-2-25
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.

更多回答(3 个)

James Tursa
James Tursa 2020-2-25
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);

hamza
hamza 2022-6-24
编辑:Image Analyst 2022-6-24
Plot the contour plots of the circles x^2+y^2 of radius 1,2, 1.41,1.73.
  1 个评论
Image Analyst
Image Analyst 2022-6-24
radii = [1, 2, 1.41, 1.73];
viscircles([zeros(4,1), zeros(4,1)], radii);
axis equal
grid on;

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Steven Lord
Steven Lord 2022-6-24
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
  2 个评论
Image Analyst
Image Analyst 2022-6-24
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
Steven Lord 2022-6-24
Note that viscircles is part of Image Processing Toolbox which means that not all users would have access to it.

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