Getting an array of points along an arc

How can I replicate the following code, which plots an arc given 2 endpoints and the radius, to instead output an array of points? (Which I can then plot using plot(x,y))

i.e. I want an array of points that will give me an arc, given two endpoints a and b, and a radius r.

figure;
a = [124.0102   62.2260];
b = [33.3602   77.9434];
r=250;
syms x y
[x,y]=solve((x-a(1))^2+(y-a(2))^2==r^2,(x-b(1))^2+(y-b(2))^2==r^2,x,y);
%plot arc
syms X Y
fig = ezplot((X-x(2))^2+(Y-y(2))^2==r^2,[min(a(1),b(1)),max(a(1),b(1)), ...
    min(a(2),b(2)),max(a(2),b(2))]); 
set(fig,'color','black','LineStyle', '--','LineWidth',2)

[KSSV]

Did you refer these: https://in.mathworks.com/matlabcentral/answers/367126-plot-an-arc-on-a-2d-grid-by-given-radius-and-end-points

https://in.mathworks.com/matlabcentral/answers/327895-how-do-i-plot-a-circular-arc-with-given-two-endpoints-and-radius

[MJ]

Yes, the code I have is based on those posts.

Answer 1

[KSSV]

clc; clear all ;
figure;
a = [124.0102   62.2260];
b = [33.3602   77.9434];
r=250;
syms x y
[x,y]=solve((x-a(1))^2+(y-a(2))^2==r^2,(x-b(1))^2+(y-b(2))^2==r^2,x,y);

x = double(x) ; 
y = double(y) ; 

xi = linspace(b(1),a(1)) ; 
yi = linspace(a(2),b(2)) ; 
[X1,Y1] = meshgrid(xi,yi) ; 
R = (X1-x(2)).^2+(Y1-y(2)).^2 ; 
idx = abs(R-r^2)<15 ; 


Xa = X1(idx) ; Ya = Y1(idx) ; 
[Xa,idx]  = sort(Xa) ; 
Ya = Ya(idx) ; 
 plot(Xa,Ya,'-*b')

Answer 2

[DGM]

This is basically the same as this answer, but maybe the explanation helps a bit:

https://www.mathworks.com/matlabcentral/answers/327895-how-do-i-plot-a-circular-arc-with-given-two-endpoints-and-radius#answer_272201

a = [124.0102 62.2260]; % endpoints [x y]
b = [33.3602 77.9434];
r = 250; % radius
npoints = 100; % number of points to draw

% Finding the coordinates of the centers of circles
p = [a(:) b(:)];
c = sym('c',[2,1],'real');
eqs = [1,1]*(p - repmat(c(:),1,2)).^2 - r^2;
[cx cy] = vpasolve(eqs,c);
C = double([cx cy]); % solutions for centers [cx(:) cy(:)]

% each set of points with distance < 2*r
% will yield two solutions for the center coordinates
% each solution describes a circle, for which there are two possible arcs
% the arc angles are conjugate and inequal; we usually want the shorter one

% draw both short arcs and their corresponding centers
for ka = 1:size(C,1)
    % find the shortest arc for this solution
    v = C(ka,:);
    p1 = p - v(:);
    alp = atand(p1(2,:)./p1(1,:));
    alp = alp + 180*(p1(1,:) < 0 & p1(2,:) > 0) - 180*(p1(1,:) < 0 & p1(2,:) < 0);
    %alp = sort(alp); % only matters if you care about parameter direction
    
    % generate points
    phi = linspace(alp(1),alp(2),npoints)';
    x = r*cosd(phi) + v(1);
    y = r*sind(phi) + v(2);
    
    % plot the short arc and center for this solution
    plot(x,y); hold on
    plot(v(1),v(2),'ok');
end
grid on
axis equal