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**Failure** Let's assume that $\displaystyle k:\; (x-u)^2+(y-v)^2=16$ is the equation of the circle with center $\displaystyle C(u,v)$ and radius 4.

Since (-2,3) and (2,-1) lie on this circle, you get two equations for the coordinates u, v of the center, namely

$\displaystyle (-2-u)^2+(3-v)^2=16$ and $\displaystyle (2-u)^2+(-1-v)^2=16$.

If you now subtract the second equation from the first, you get $\displaystyle 8u-8v+8=0$, and hence $\displaystyle u=v-1$.

Replacing $\displaystyle u$ in the first equation by $\displaystyle v-1$ gives you the quadratic equation $\displaystyle (-1-v)^2+(3-v)^2=16$.

Now solve for $\displaystyle v$: this gives you two solutions $\displaystyle v_1,v_2$ for $\displaystyle v$. Finally, pair these two solutions with the corresponding x-coordinates of C, namely $\displaystyle u_{1,2}=v_{1,2}-1$, and you are done.