# How do I plot a circular arc with given two endpoints and radius?

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Munish Kumar on 3 Mar 2017
Edited: John D'Errico on 28 Jun 2017
I have two points P1 and P2 with coordinates (x1,y1) & (x2,y2), respectively. I need to plot a circular arc of a given radius 'r' passing through P1 and P2.
Equation of circle (x-a)^2 + (y-b)^2 = r^2 was tried which yielded a semicircle. Then I tried the parametric equation of circle x = a + r*cos(t), y = b + r*sin(t), which again yielded a complete circle.
Is there any way to draw a circular arc through P1 & P2?
John D'Errico on 28 Jun 2017
Note that there are always two (really 4) possible arcs, as long as the radius is at least twice the distance between the two points. So the solution will not be unique. Assuming you want the smallest possible arc, that reduces it always 2 arcs if the problem is solvable at all, and the radius isfinite. If the radius approaches inf, then the two arcs both approach the straight line between the points as a limit.

Andrei Bobrov on 28 Jun 2017
p = [4 10;7 4]; % p - this is our points p1 & p2, p =[x1,x2;y1,y2];
r = 7; % r - radius
% Finding the coordinates of the centers of circles
% xy = [x1, y1; x2, y2]
a = sym('a',[2,1],'real');
eqs = [1,1]*(p - repmat(a(:),1,2)).^2 - r^2;
sol = vpasolve(eqs,a);
ss = struct2cell(sol);
xy = double([ss{:}]);
% example: The arc of a circle with center at xy(1,:)
v = xy(1,:);
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);
asort = sort(alp);
phi = linspace(asort(1),asort(2),100)';
figure
plot(r*cosd(phi) + v(1),r*sind(phi) + v(2),'-b',v(1),v(2),'ok')
grid on

jack wu on 28 Jun 2017
Edited: John D'Errico on 28 Jun 2017
%first input
a=[0 1]; %P1
b=[1 0]; %P2
%next solution
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
ezplot((X-x(1))^2+(Y-y(1))^2==r^2,[min(a(1),b(1)),max(a(1),b(1)), ...
min(a(2),b(2)),max(a(2),b(2))])
axis equal
figure
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))])
axis equal
John D'Errico on 28 Jun 2017