<s>[INST] System of equations with complex numbers-circles The system of equations
\begin{align*}
|z - 2 - 2i| &= \sqrt{23}, \\
|z - 8 - 5i| &= \sqrt{38}
\end{align*}
has two solutions $z_1$ and $z_2$ in complex numbers. Find $(z_1 + z_2)/2$.
So far I have gotten the two original equations to equations of circles,
$(a-2)^2 +(b-2)^2=23$ and $(a-8)^2+(b-5)^2=38$.
From here how do I find the solutions?
Thanks.
 [/INST] You're given the distances of the solution points to $A = 2+2i$ and $B = 8+5i.$
One of the solutions and these two complex numbers give you a triangle in the complex plane, and you know the lengths of all the sides.
We can use Law of Cosines to find the angle $\theta$ that has $2+2i$ as its vertex:
$$38 = 45 + 23 - 2 \sqrt{45} \sqrt{23} \cos \theta \to \cos \theta \approx. 0.466252.$$
The projection of $AZ_1$ onto $AB$ is then $\sqrt{23} \cdot 0.466252 \approx 2.23606$.
Why calculate this?  Because the other solution is symmetric on the other side of $AB$, and what you're asked to find, in essence, is the average of the two solutions, which lies on $AB$.
The unit vector from $A$ to $B$ is $(6 + 3i)/\sqrt{45} = (2+i)/\sqrt{5}.$
So, the solution is
$$(Z_1 + Z_2)/2 = 2 + 2i + 2.23606 \cdot (6 + 3i)/\sqrt{45} \approx 4 + 3i.$$
(The fact that this worked out so nicely makes me think there was an easier way, but oh well ...) 
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