## Plotting multiple objects on same axis

### Jide Williams (view profile)

on 26 Oct 2018
Latest activity Commented on by Jide Williams

on 26 Oct 2018

### Kevin Chng (view profile)

I am new to MATLAB but I am trying to plot multiple circles on the same axis. I am using rand function to generate the centers because I want some of them to overlap, but it ends up plotting just one circle. Below is my code;
r = 0.005; x = rand(100,1); y = rand(100,1); th = 0:pi/100:2*pi; for i = 1:length(x) & 1:length(y) a = r*cos(th) + x(i); b = r*sin(th) + y(i); axis tight; hold on plot(a,b) end

### Kevin Chng (view profile)

on 26 Oct 2018
Edited by Kevin Chng

### Kevin Chng (view profile)

on 26 Oct 2018

r = 0.005;
x = rand(100,1);
y = rand(100,1);
th = 0:pi/100:2*pi;
for i = 1:1:length(x)
a = r*cos(th) + x(i);
b = r*sin(th) + y(i);
figure(1)
axis tight;
hold on
plot(a,b)
pause(0.1)
end
Slight change your code, it is very interesting, if you add pause(), then you see the animation. However, you delete the pause() if you don't want it.

on 26 Oct 2018

on 26 Oct 2018

r = 0.005;
x = rand(100,1);
y = rand(100,1);
th = 0:pi/100:2*pi;
a=zeros(1,numel(x)) % preallocation for speed and efficiency
b= zeros(1,numel(y))
for i = 1:numel(x)
a(i) = r*cos(th(i)) + x(i);
b(i) = r*sin(th(i)) + y(i);
end
axis tight;
hold on
plot(a,b,'ob')

on 26 Oct 2018
the actual question was to plot a point of circles centre point and not the radius of the circle as far as I understood
jonas

### jonas (view profile)

on 26 Oct 2018
Yep, you are entirely correct in that it's a poorly written question. One of the things you learn on this forum is filling in blanks and interpreting ambiguities, which is actually quite useful when teaching :)
With that said, the notations r and the values of th (0 to 2pi) and the fact that OP accepted @Keving Chng's answer would imply that OP intended the radius to go as input.

on 26 Oct 2018
Wow thanks Jonas completely agree with you , one thing experience is gained day by day :)

### jonas (view profile)

on 26 Oct 2018

If you have the image processing toolbox available, then there is a one-line solution to your problem.
viscircles([x,y],ones(size(x)).*r)

jonas

### jonas (view profile)

on 26 Oct 2018
r is a scalar, but the input to the function must be a vector of the same size as x and y. That's why you simply multiply it by a number of ones, like:
1 0.005
1 *. 0.005 = 0.005
1 0.005
... ...
jonas

### jonas (view profile)

on 26 Oct 2018
Viscircles basically does the same thing as your code, but probably more effeciently. I have not actually timed and compared.
Jide Williams

on 26 Oct 2018
Thanks again