<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] We can also treat this in a way that has little to do with complex numbers.  The line connecting the centers of the circles at $ \ (2, \ 2) \ $ and $ \ (8, \ 5 ) \ $ has a slope of $ \ \frac{5 - 2}{8 - 2} \ = \ \frac{1}{2} \ $ and has the equation
$$ y \ - \ 2 \ = \ \frac{1}{2} \ (x - 2) \ \ \Rightarrow \ \ y \ = \ \frac{1}{2} \ x \ + \ 1 \ \ . $$
Using a theorem from classical geometry, the two intersection points of the two circles lie on a mutual chord, of which the line connecting the centers of the circles is its perpendicular bisector (a fact also used in some way by M. Wind and John).  Thus, we expect the chord to have a slope of $ \ -2 \ $ .
A relation for the coordinates of the intersection points of the circles can be found by subtracting one circle equation from the other; thus,
$$ \ x^2 \ - \ 16x \ + \ 64 \ + \ y^2 \ - \ 10y \ + \ 25 \ = \ 38 $$
$$ - \ ( \ x^2 \ - \ 4x \ + \ 4 \ + \ y^2 \ - \ 4y \ + \ 4 \ = \ 23 \ ) $$
$$ -------------------- $$
$$ -12x \ + \ 60 \ - \ 6y \ + \ 21 \ = \ 15 $$
$$ \Rightarrow \ \ y \ = \ -2x \ + \ 11 \ \ . $$
The chord and its perpendicular bisector meet at
$$ \frac{1}{2} \ x \ + \ 1 \ = \ -2x \ + \ 11 \ \ \Rightarrow \ \ \frac{5}{2} \ x \  =  \ 10 \ \ \Rightarrow \ \ x \ = \ 4 \ \ \Rightarrow \ \ y \ = \ 3 \ \ , $$
found by inserting the value for  $ \ x \ $  into either of the linear equations we've determined.  Hence, the midpoint of the chord containing the intersections of the circles lies at $ \ z \ = \ 4 \ + \ 3i \ $ in the complex plane.
Though we aren't asked to find them, we can also locate $ \ z_1 \ \ \text{and} \ \ z_2 \ $ . Inserting $ \  y \ = \ -2x \ + \ 11 \ $ into the first circle equation, we produce
$$ (x - 2)^2 \ + \ (-2x + 11 - 2)^2 \ = \ 23 \ \ \Rightarrow \ \ 5x^2 \ - \ 40x \ + \ 62 \ = \ 0 \ \ , $$
for which the roots are $ \ x \ = \ 4 \ \pm \ \frac{3}{5} \ \sqrt{10} \ $ . [Using the other circle equation instead yields exactly the same quadratic equation.]
We may then calculate $ \ y \ = \ -2 \ (4 \ \pm \ \frac{3}{5} \ \sqrt{10}) \ + \ 11 \ = \ 3 \ \mp \ \frac{6}{5} \ \sqrt{10} \ $ .  Thus, we find
$$ z_1 \ = \ (4 \ - \ \frac{3}{5} \ \sqrt{10}, \  3 \ + \ \frac{6}{5} \ \sqrt{10}) \ \ , \ \ z_2 \ = \ (4 \ + \ \frac{3}{5} \ \sqrt{10}, \  3 \ - \ \frac{6}{5} \ \sqrt{10}) \ \ . $$
(Some brief work with right triangles then tells us that the length of the chord between $ \ z_1 \ \ \text{and} \ \ z_2 \ $ is $ \ \sqrt{\frac{36 + 144}{25} \ \cdot \ 10} \ = \ \sqrt{72} \ = \ 6 \ \sqrt{2} \ $ .)
</s>