There are three equations.
$$\sqrt{{(X_{1}-X_{0})^2} +{(Y_{1}-Y_{0})^2}} = AC$$
$$\sqrt{{(X_{2}-X_{0})^2} +{(Y_{2}-Y_{0})^2}} = BC$$
$$\sqrt{{(X_{0}^2+Y_{0}^2)}} = CO$$
If we assume that $$A(X_{2},Y_{2})$$
and $$B(X_{2},Y_{2})$$ are the location of 2 points. And $$C(X_{0},Y_{0})$$ is our point of reference. CO is the distance between point of reference and origin.
Assuming this information solve for $$X_{0} and Y_{0}$$
I'm not sure if this is possible.
On
If you want to check if it is feasible, rewrite the equations as $$\Delta_1={(X_{1}-X_{0})^2} +{(Y_{1}-Y_{0})^2} -AC^2\tag 1$$ $$\Delta_2={(X_{2}-X_{0})^2} +{(Y_{2}-Y_{0})^2} - BC^2\tag 2$$ $$\Delta_3=X_{0}^2 +Y_{0}^2 - CO^2\tag 3$$
Assume that $\Delta_1=\Delta_2=\Delta_3=0$ and develop $(\Delta_1-\Delta_3)$ and $(\Delta_2-\Delta_3)$; this gives two linear equations in $(X_0,Y_0)$ which are easy to solve. Using them, recompute the values of $\Delta_1,\Delta_2,\Delta_3$. If they are all $0$, the solution is obtained. Otherwise, no solution.
What you also could do is to consider the function $$F(X_0,Y_0)=\Delta_1^2+\Delta_2^2+\Delta_3^2$$ and minimize it with respect to $X_0,Y_0$. If its minimum value is $0$, then you have the solution. The starting point for the minimization would be the one obtained by the first step.
The circle of radius $AB$ centred at $(X_1, Y_1)$ and the circle of radius $BC$ centred at $(X_2,Y_2)$ will intersect in $0$, $1$ or $2$ points. "Generically" none of these will be on the circle of radius $CO$ centred at the origin.