<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] The first thing to do is to make a drawing of the two circles. Label their centres $C_1$ and $C_2$, and the two intersections of the circles $I_1$ and $I_2$. Now connect these four points by straight lines, and label as $P$ the point half-way between $I_1$ and $I_2$, which is on the line connecting $C_1$ and $C_2$. 
We recognize that we have four triangles, all with a $90$ degree angle. So we can apply Pythagoras. For convenience we label the line-piece $C_1$ to $P$ as $a$; $C_2$ to $P$ as $b$; $I_1$ (or $I_2$) to $P$ as $c$. Now: 
$$(a+b)^2 = 45$$
$$a^2 + c^2 = 23$$
$$b^2 + c^2 = 38$$
We eliminate $c^2$ by subtracting the second equation from the third, yielding:
$$(b-a)(b+a) = 15$$
Dividing the first equation by this result gives
$$ \frac {b+a}{b-a} = 3$$
From which it follows that $b = 2a$. So $a$ is one-third of the distance between $C_1$ and $C_2$. Therefore the position of point $P$ is given by $2+2i$ + $(6+3i)/3$ = $4+3i$.  
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