The point $(x_0,y_0)$ is on three circles, with equations
$$C_i(x,y)=0, \qquad i=1,2,3,$$
For a point $(x,y)$ to be on the first two circles, we must have $C_1(x,y)=C_2(x,y)$. Expand and simplify. We get a linear equation.
Similarly, we have $C_2(x,y)=C_3(x,y)$. Expand and simplify. We get another linear equation.
We now have a system of two linear equations in two unknowns. Solve for $x$ and $y$.
Note that bad things can happen. If we put down points and distances "at random" it is quite likely that there will be no point $(x_0,y_0)$ that satisfies your conditions. In certain fairly rare cases, there may be more than one point that satisfies your conditions.
Details: The point $(x_0,y_0)$ is distance $d_1$ from $(x_1,y_1)$. So $(x_0,y_0)$ is on the circle with centre $(x_1,y_1)$ and radius $d_1$. This circle has equation
This equation expands to
Similarly, $(x_0,y_0)$ lies on the circle with equation
When we set the two left-hand sides equal to each other, the $x^2$ and $y^2$ terms cancel, and we arrive at the linear equation
$$2(x_2-x_1)x +2(y_2-y_1)y +x_1^2-x_2^2+y_1^-y_2^2 -d_1^2+d_2^2=0.$$
This is a linear equation in $x$ and $y$. When the two circles intersect in two points, it is the equation of the line through these two points.
In the same, we obtain a second linear equation in $x$ and $y$, that is, the equation of (we hope) another line. Then $(x_0,y_0)$ lies on both lines, so can be easily found.
It is possible that the two lines will be the same. This happens when the points $(x_i,y_i)$ all lie in one line (are collinear). In that case, all is not lost. Use one of our linear equations to solve for $y$ in terms of $x$, and substitute in the equation of one of our circles. After a while, we get a quadratic equation in $x$. Solve. In general there will be two solutions, and thus two possibilities for $(x_0,y_0)$, symmetric about the line that contains our points $(x_i,y_i)$ ($i=1,2,3$). A sketch shows that we cannot do any better, for when our three points are collinear, there are in general two possibilities for $(x_0,y_0)$.