# Plot an Arc on a 2D Grid by given radius and end points

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TS Low on 15 Nov 2017
I have one question, how do I plot the arc on a graph by giving the radius and it end points? It start points is the points set by me take example (2,2). I need draw an arc with radius 3 and end point (5,5) How to write the code for this

Roger Stafford on 15 Nov 2017
Edited: Roger Stafford on 15 Nov 2017
Point vectors A and B must be column vectors
A = randn(2,1); % Point A to be on circle circumference
B = randn(2,1); % Same with point B
d = norm(B-A);
R = d/2+rand; % Choose R radius >= d/2
C = (B+A)/2+sqrt(R^2-d^2/4)/d*[0,-1;1,0]*(B-A); % Center of circle
a = atan2(A(2)-C(2),A(1)-C(1));
b = atan2(B(2)-C(2),B(1)-C(1));
b = mod(b-a,2*pi)+a; % Ensure that arc moves counterclockwise
t = linspace(a,b,1000);
x = C(1)+R*cos(t);
y = C(2)+R*sin(t);
plot(x,y,'y-',C(1),C(2),'w*')
axis equal
Note that another possible center can be obtained by
C2 = (B+A)/2-sqrt(R^2-d^2/4)/d*[0,-1;1,0]*(B-A);
Note 2: If C is chosen, the arc will be <= pi. If C2 is used, arc will be >= pi
TS Low on 19 Nov 2017 Where is the point A? in term of X1 and Y1
A = randn(2,1);?
I didn use your solution because i replace the value with other but cant solve.

KSSV on 15 Nov 2017

Image Analyst on 19 Nov 2017
TS Low on 19 Nov 2017
yes, it is an arc
by i need prompt user for start point end point and radius
so this might be no work for me

%Equation of a circle with centre (a,b) is (x-a)^2+ (y-b)^2 = r^2
%Circle Centre (1,1), radius = 10
k=1; %counter
c =1 ; % value of x at the centre of the circle
while c <=11
x(k) = c ;
vv = (c-1)^2 ;
y (k) = 1 + real (sqrt (100 - vv) );
c= c + 0.02;
k=k+1;
end
plot (x, y, 'r')
axis equal Navinda Wickramasinghe on 17 Sep 2020
The solution was perfect. As Roger mentions, just make sure to provide the endpoint coordinates in the counterclockwise direction.