How can I solve three unknowns with three equations in Matlab?

I am new to Matlab and trying to solve these equations in Matlab.

(1.92-x)^2+(-0.88-y)^2=r^2

(1.42-x)^2+(1.22-y)^2=(r+343(0.00638))^2

(-1.51-x)^2+(1-y)^2=(r+343(0.01149))^2

I am very unsure of how this can be done. This must be possible in a software like Matlab. What code do you think it is best to use when solving this?

[Christian Klaue]

Gauss elimination method

[John D'Errico]

NO. These are not linear equations. Gaussian elimination does not handle quadratic equations.

Answer 1

[John D'Errico]

Easily enough done using the symbolic toolbox.

syms x y r

E1 = (1.92-x)^2+(-0.88-y)^2 == r^2;
E2 = (1.42-x)^2+(1.22-y)^2 == (r+343*(0.00638))^2;
E3 = (-1.51-x)^2+(1-y)^2 == (r+343*(0.01149))^2;

result = solve(E1,E2,E3)
result = 
    r: [2x1 sym]
    x: [2x1 sym]
    y: [2x1 sym]

vpa(result.r)
ans =
 -0.015481067535873645827800666354194
  -0.30809941998244195263749288776231

vpa(result.x)
ans =
 1.9293936680622001322345796823646
 1.7349298832558521137294030854302

vpa(result.y)
ans =
 -0.89230538306545075129564568787526
 -0.63367845307033202264344475680987

So there are two possible solutions.

These are the equations of three circles, with fixed centers, and variable radii that depends on r. See that r came out negative in all three cases, but that is irrelevant. The radii of the circles are not negative. (With only a little effort, the three circles can be plotted, along with the solutions.)

Again, Gaussian elimination is not possible here, although one could solve the problem using pencil and paper.

[Elias Andersen]

It Outputs:

Error in eqn_3ukn (line 2)

syms x y r

Why does this come up as the error?

Answer 2

[Disha]

syms x y r E1 = (1.92-x)^2+(-0.88-y)^2 == r^2; E2 = (1.42-x)^2+(1.22-y)^2 == (r+343*(0.00638))^2; E3 = (-1.51-x)^2+(1-y)^2 == (r+343*(0.01149))^2; result = solve(E1,E2,E3) result = r: [2x1 sym] x: [2x1 sym] y: [2x1 sym]