How can I plot a simple circle not centered at the origin? example: (x-5)^2+(y-10)^2=4

example: (x-5)^2+(y-10)^2=4 I actually want to plot several overlapping circles on the same graph. I have never used matlab prior to this week so if there is a way to do this using SIMPLE function, plot, ezplot, etc. that's what I'm looking for. Thank you.

Answer 1

[Amit]

You make a function for plotting the circle, something like this:

EDIT:

function plotcircle(x1,y1,r,c)
% x1,y1: center
% r = radius
theta = 0:0.05:2*pi;
hold on;
plot(x1+r*cos(theta),y1+r*sin(theta),c);
axis equal;

For for every circle, you can call this function: like

plotcircle(5,10,2,'r');

[cassie]

That's exactly what i was looking for. I plotted 5 circles, but i want to change the colors and plotcircle(5,10,2,'r'); for example doesn't work. can you tell me what I'm missing?

[Amit]

See the edited code. The prior function didn't have the color variable. Now it does.

[cassie]

I tried it just like you have it here by changing the function to include the "c" but it throws the following error: Error using plotcircle Too many input arguments.

Error in Project1 (line 3) plotcircle(5,10,4,'y');

[Amit]

Did you tried the edited code (not just the edited function call).

[cassie]

I did.... plot(x+r*cos(theta),y+r*sin(theta),c)

[cassie]

I found my mistake... I didn't add the ,c at the very top. Thank you so much for your help!

Answer 2

[Image Analyst]

Code samples are in the FAQ: http://matlab.wikia.com/wiki/FAQ#How_do_I_create_a_circle.3F

[cassie]

I looked there but didn't find exactly what I was looking for. Thank you though.

[Image Analyst]

You'll notice that the code there:

xCenter = 12;
yCenter = 10;
theta = 0 : 0.01 : 2*pi;
radius = 5;
x = radius * cos(theta) + xCenter;
y = radius * sin(theta) + yCenter;
plot(x, y);
axis square;
xlim([0 20]);
ylim([0 20]);
grid on;

is virtually identical to Amit's code except he named the variables differently and put the equations into the plot() function instead of into variables (where they can be accessed later if desired). So it seemed exactly what you needed, to me at least.

[Amit]

Image Analyst, I did not copy the code from the wikia page. The simplest way get all the cartesian coordinate values for a circle is the parametric form equation, which is what I used.

[Image Analyst]

I didn't think you did. It's so very simple so I'm sure you thought of it independently. I just brought it up because I'm hoping the poster, or others, might also scan the other interesting stuff in there. A lot of people don't know about the existence of the FAQ, and won't unless we keep plugging it now and then.

[Amit]

I am really sorry then. I thought by 'he named the variable differently' you meant I copied the code.