How to change the ratio of sectors for a circle with 8 equal sectors

I have a circle with 8 equal sectors and radius one(Below is the code I used)

Now, I want to be able to move the circle around whilst keeping the partition lines in place(like in the image below), How can I do this but also be able to calculate the area of each sector for the new circle ?

[Walter Roberson]

It is better to post the code instead of an image of the code. There are no existing MATLAB versions that are able to execute images of code.

[John D'Errico]

Remember that you want a fast answer. If you post only a picture of your code, anyone who wants to help you needs to type it in, from the picture. Since you could trivially have pasted the code directly in as text, why not do so? Then it is far easier for someone to help you. And why would you possibly want to make it more difficult to get help?

[Torsten]

Movement of the circle is arbitrary or does it suffice to consider horizontal movement ?

[Matt J]

I want to be able to move the circle around whilst keeping the partition lines in place(like in the image below), How can I do this but also be able to calculate the area of each sector for the new circle ?

Why would the movement of the circle change the area of the sectors?

Do you mean you the center of the radiating pattern is moving while the circumference is not? If so, those aren't really "sectors" anymore, strictly speaking.

[Walter Roberson]

@Matt J The center point that is acting as the intersection, is moving. It is obvious that the areas of the 8 sectors are going to change.

Answer 1

[Matt J]

One possibility:

dTheta=360/8;  %sector angle
t=linspace(0,dTheta)';

sector=polyshape([0,0;cosd(t),sind(t)]);
circle=arrayfun(@(z)rotate(sector,z),(0:7)*dTheta);

newcircle=newCenter(circle, [+0.5,0]);

sectorAreas=area(newcircle)

sectorAreas = 1×8

0.8012 0.4625 0.1979 0.1092 0.1092 0.1979 0.4625 0.8012

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plot(newcircle); axis equal

●

function circle=newCenter(circle, t)

 R=sqrt(sum(area(circle))/pi); %radius
 
 mag=1+norm(t)/R;
 
 Circle=scale(circle,mag,[0,0]);

 circle=intersect( translate(union(circle),t)  ,  Circle);

end

Answer 2

[DGM]

Alternative interpretation, also using polyshape():

This still subdivides the outer circle into equal arc lengths. I'm not really sure if that is intended.

The circle and line intersection positions can be defined independently, so you can choose how you want to configure it.

nsectors = 8; % number of sectors
lineint = [0 0.2]; % position of line intersection
center = [0.5 -0.2]; % position of circle center
radius = 1; % radius of outer circle
np = 100; % points per arc

th = linspace(0,2*pi,nsectors+1); % breakpoint angles
P = repmat(polyshape(),[nsectors 1]);
for k = 1:nsectors
	arcth = linspace(th(k),th(k+1),np).';
	V = [lineint; [cos(arcth) sin(arcth)] + center];
	P(k) = polyshape(V);
end

% i'm using different colors since this is slightly longer than lines()
% this will avoid repeated colors and clearly indicates the sequence order
colororder(parula(nsectors))
plot(P); axis equal; grid on

area(P)

ans = 8×1

0.5109 0.3245 0.1781 0.1573 0.2745 0.4609 0.6073 0.6281

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[Torsten]

Cited from the OP:

Now, I want to be able to move the circle around whilst keeping the partition lines in place(like in the image below),

So I think @Matt J 's interpretation is correct.

[DGM]

I saw that, but given the manner in which the original code constructed the plot and the crudeness of the drawing, I considered it ambiguous enough that it would be worth throwing down an example anyway. It's my time to waste, after all.