Following up on the comment by user141267: an efficient way to give more or less weight to equations in an overdetermined system is to rescale them; that is, multiply both $A$ and $b$ by a diagonal matrix $W$ on the left. Here is an example:
$$\begin{cases} x+ y & =10 \\ x+2y &= 14 \\ x+3y &= 40\end{cases}$$
To solve this system using least squares, I used lsq(A,b) in Scilab with
$$A = \begin{pmatrix} 1 & 1 \\ 1 & 2 \\ 1 & 3 \end{pmatrix}, \quad b = \begin{pmatrix} 10 \\ 14 \\ 30 \end{pmatrix}$$
and got $x=-2 $, $y=10$. The right hand side for this solution is $(8, 18, 28)^T$.
But suppose the first equation is very important / certain, while the last one is the least important. If I let $W$ be the diagonal matrix with entries $(4,1,1/2)$ and run lsq(W*A,W*b), the result is $x= 2.84$, $y=7.07$. The right hand side for this solution is $(9.91, 16.98, 24.05)^T$. So, the first equation is satisfied almost exactly, while the last equation is pretty far from target.
Rewrite the system of equations in their original form,
$$(x-x_0)^2+(y-y_0)^2 =r^2\tag{1}$$
$$(x-x_1)^2+(x-y_1)^2 = (r_1+r)^2\tag{2}$$
$$(x-x_2)^2+(x-y_2)^2 = (r_2+r)^2\tag{3}$$
where $r$ is the radius of the circle with the center $(x,y)$ to be solved.
As seen below, $x$ and $y$ depend on $r$ linearly. It is more convenient to maintain symmetry by solving for $r$ first and then for $(x,y)$.
From (2)-(1) and (3)-(1), we get a pair of linear equations,
$$a_1x+b_1y=c_1+2r_1r\tag{4}$$
$$a_2x+b_2y=c_2+2r_2r\tag{5}$$
where the coefficients are all known and given by,
$$a_1=2(x_0-x_1), \>\>\>\> b_1=2(y_0-y_1),\>\>\>\>\> c_1=r_1^2+x_0^2+y_0^2-x_1^2-y_1^2$$
$$a_2=2(x_0-x_2), \>\>\>\> b_2=2(y_0-y_2),\>\>\>\> c_2=r_2^2+x_0^2+y_0^2-x_2^2-y_2^2$$
Solve the linear equations (4) and (5) for $x$ and $y$ in terms of $r$,
$$x=d_1r+e_1,\>\>\>\>\> y = d_2r+e_2\tag{6}$$
where the known coefficients are,
$$d_1 = \frac{2r_1b_2-2r_2b_1}{a_1b_2-a_2b_1},\>\>\>\>\>\>e_1 = \frac{c_1b_2-c_2b_1}{a_1b_2-a_2b_1} $$
$$d_2 = \frac{2r_1a_2-2r_2a_1}{a_2b_1-a_1b_2},\>\>\>\>\>\>e_2 = \frac{c_1a_2-c_2a_1}{a_2b_1-a_1b_2} $$
Plug (6) back into (1) to get a quadratic equation in $r$,
$$ar^2+br+c=0\tag{7}$$
with the coefficients given by,
$$a=d_1^2+d_2^2-1,\>\>\> b= 2(e_1-x_0)d_1+2(e_2-y_0)d_2,\>\>\>c=(e_1-x_0)^2+(e_2-y_0)^2$$
The equation (7) for $r$ can be solved with the standard quadratic formula,
$$r = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
Afterwards, plug $r$ into (6) to obtain the solutions for $x$ and $y$.
Best Answer
Your system is described by the augmented matrix $$ A= \left[\begin{array}{rrrr|r} 0 & 1 & 1 & 0 & r_{1} \\ 0 & 1 & 0 & 1 & c_{1} \\ 1 & 1 & 0 & 0 & d_{1} \\ 0 & 0 & 1 & 1 & d_{2} \\ 1 & 0 & 1 & 0 & c_{2} \\ 1 & 0 & 0 & 1 & r_{2} \end{array}\right] $$ Row-reducing the system gives $$ \DeclareMathOperator{rref}{rref}\rref A= \left[\begin{array}{rrrr|r} 1 & 0 & 0 & 0 & -\frac{1}{2} \, c_{1} + d_{1} + \frac{1}{2} \, d_{2} - \frac{1}{2} \, r_{1} \\ 0 & 1 & 0 & 0 & \frac{1}{2} \, c_{1} - \frac{1}{2} \, d_{2} + \frac{1}{2} \, r_{1} \\ 0 & 0 & 1 & 0 & -\frac{1}{2} \, c_{1} + \frac{1}{2} \, d_{2} + \frac{1}{2} \, r_{1} \\ 0 & 0 & 0 & 1 & \frac{1}{2} \, c_{1} + \frac{1}{2} \, d_{2} - \frac{1}{2} \, r_{1} \\ 0 & 0 & 0 & 0 & c_{1} + c_{2} - d_{1} - d_{2} \\ 0 & 0 & 0 & 0 & -d_{1} - d_{2} + r_{1} + r_{2} \end{array}\right] $$ This implies that your system is solvable if and only if \begin{align*} c_1+c_2-d_1-d_2 &= 0 \\ -d_1-d_2+r_1+r_2 &= 0 \end{align*} If these conditions are satisfied, then your system is solved by \begin{align*} w &=-\frac{1}{2} \, c_{1} + d_{1} + \frac{1}{2} \, d_{2} - \frac{1}{2} \, r_{1} \\ x &= \frac{1}{2} \, c_{1} - \frac{1}{2} \, d_{2} + \frac{1}{2} \, r_{1}\\ y &= -\frac{1}{2} \, c_{1} + \frac{1}{2} \, d_{2} + \frac{1}{2} \, r_{1}\\ z &=\frac{1}{2} \, c_{1} + \frac{1}{2} \, d_{2} - \frac{1}{2} \, r_{1} \end{align*}