A function to draw a circle with only radius as parameter?

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Peter
Peter 2012 年 11 月 12 日
Hello I need help to create a function that will help me draw a circle with a given radius.
So the function can only have the radius as parameter.
For example if I type draw_circle(5) I want to have a plot of a circle with radius 5. Since the equation is y^2+x^2=r^2 ---> y=+/-sqrt(r^2-x^2) it feels like I need two parameters in the function one for x and one for r.
So basically can someone help me construct a function draw_circle(r) that plots a circle with radius 5.
Thanks!

採用された回答

Azzi Abdelmalek
Azzi Abdelmalek 2012 年 11 月 12 日
編集済み: Azzi Abdelmalek 2012 年 11 月 12 日
function cercle(r)
x=-r:.01:r
y=sqrt(r^2-x.^2)
hold on;
plot(x,y);
plot(x,-y)
call the function
cercle(5)
  3 件のコメント
Azzi Abdelmalek
Azzi Abdelmalek 2012 年 11 月 12 日
編集済み: Azzi Abdelmalek 2012 年 11 月 12 日
What circle(5) is doing inside a function?
save the function I writted above, it does'nt contain circle(5) . then in matlab command type
circle(5)
Peter
Peter 2012 年 11 月 13 日
Sorry I have been programming in python too much. Thank you it worked.

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その他の回答 (2 件)

Matt Fig
Matt Fig 2012 年 11 月 12 日
編集済み: Matt Fig 2012 年 11 月 12 日
function [] = plot_circle(R)
% Plots a circle of radius R
[X,Y] = cylinder(R,1000);
plot(X(1,:),Y(1,:))
axis equal
Save this as plot_circle.m then call from the command line:
plot_circle(4) % Radius 4
If you just want a function handle that can do it, try this one:
f = @(R) plot(R*cos(0:.001:2*pi),R*sin(0:.001:2*pi));
f(3)
axis equal

Image Analyst
Image Analyst 2012 年 11 月 13 日

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