How to plot circle by one single equation?
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I need code which plot the circle in one single equation (variable). I have the code but i need code of single equation, the code which i have, it is composed of two equations as follow :
r=2;
x=0;
y=0;
th = 0:pi/6:2*pi;
xunit = r * cos(th) + x;
yunit = r * sin(th) + y;
plot(xunit, yunit);
采纳的回答
John D'Errico
2018-5-13
编辑：John D'Errico
2018-5-13
As one "equation" what does that mean? About the best I can offer is this:
r=2;
x0=0;
y0=0;
ezplot(@(x,y) (x-x0).^2 + (y-y0).^2 -r^2)
axis equal
Be careful that the units are equal for the two axes, else it ill not look circular.
I could also have done it using fimplicit.
r=2;
x0=0;
y0=0;
syms x y
fimplicit((x-x0).^2 + (y-y0).^2 -r^2)
axis equal
So, in either case, only one equation.
If you really insist on only one "variable, then you need to use a polar coordinate transformation. Thus, you can do it in terms of polar angle theta, as:
r=2;
x0=0;
y0=0;
theta = linspace(0,2*pi,100);
plot(x0 + r*cos(theta),y0 + r*sin(theta),'-')
axis equal
So only one variable, theta. If you are hoping for something else, something more or less, then you need to explain carefully what the goal is here.
Are you looking for a simple way to plot a circle? Perhaps try this:
cplot = @(r,x0,y0) plot(x0 + r*cos(linspace(0,2*pi)),y0 + r*sin(linspace(0,2*pi)),'-')
Now you have a little function that will plot a circle.
cplot(2,0,0)
hold on
cplot(3,1,2)
cplot(1,-1,1)
axis equal
2 个评论
Abdulrahman Saad
2019-10-5
Thank you very much
krishan Gopal
2021-12-9
Hi, can you tell me how to draw line pattern inside the circle
更多回答（3 个）
Aditya Gupta
2020-6-26
Adding to the MVP's (perfect) answer, I found one more way:
radius = 2;
originX = 0;
originY = 0;
rectangle('Position',[originX-radius originY-radius 2*radius 2*radius], ...
'Curvature',[1 1]);
axis equal
Harshith pothuri
2021-1-14
0 个投票
r=2;
x0=0;
y0=0;
theta = linspace(0,2*pi,100);
plot(x0 + r*cos(theta),y0 + r*sin(theta),'-')
axis equal
krishan Gopal
2021-12-9
0 个投票
Hi, can you tell me how to draw line pattern inside the circle
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