Nonlinear system Circle problem

I am supposed to determine with 3 points the midpoint and the radius of a circle

 clear all; close all; clf;

 x=[5 8 4]'; y=[9 8 3]';
X = (x(3)-x(2))/2; Y=(y(1)-y(3))/2; R= ; p=[X Y R]';
for iter=1:5
    f=[ (x(1)-X)^2+(y(1)-Y)^2-R^2
        (x(2)-X)^2+(y(2)-Y)^2-R^2
        (x(3)-X)^2+(y(3)-Y)^2-R^2 ];
    disp(norm(f))
    J=[ 2*(x(1)-X)  2*(y(1)-Y)
        2*(x(2)-X)  2*(y(2)-Y)
        2*(x(3)-X)  2*(y(3)-Y) ];
    dp=-J\f; p=p+dp;
    X=p(1); Y=p(2); R=p(3);
end, p
plot(x,y,'o')

This is my code thus far, is my X=(x(3)-x(2))/2; Y=(y(1)-y(3))/2 correct, what should 'R=' be?

Best Regards Aldo

[Torsten]

You have three equations in three unknowns.

How can the Jacobian for Newton's method be 3x2 in this case ?

Best wishes

Torsten.

[Aldo]

Yes I will change that, do you know what R= should be equal to?

[Torsten]

The value for R is just an initial guess.

If your guesses for X and Y are reasonable, you could initialize R as

sqrt((x1-X)^2+(y1-Y)^2) or
sqrt((x2-X)^2+(y2-Y)^2) or
sqrt((x3-X)^2+(y3-Y)^2)

Best wishes

Torsten.

[Aldo]

How should the Jacobian be? Shouldn't it be J_11 J_12 J_21 J_ 22. Shouldn't R be considered a constant?

Best Regards

[Torsten]

You have three equations in the unknowns X,Y and R. Thus the Jacobian is (3x3):

[df1/dX df1/dY df1/dR
 df2/dX df2/dY df2/dR
 df3/dX df3/dY df3/dR]

Best wishes

Torsten.

Answer 1

[KSSV]

http://www.regentsprep.org/regents/math/geometry/gcg6/RCir.htm