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\begin{document}

{\bf Question}

Determine all the (complex) values of $s$ for which the four
points $1 +i$, $2i$, $-3-2i$, and $s$ lie on a circle in
$\overline{\bf C}$.
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{\bf Answer}

We use the cross ratio. Specifically the cross ratio is real if
and only if the 4 points lie on a circle in $\bf{C}$:

$f(1+i,2i,-3-2i,s)$

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$=\ds\frac{1+i-s}{1+i-2i} \cdot \ds\frac{-3-2i-2i}{-3-2i-s}$ write
$s=x+iy$

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$=\ds\frac{1-x+i(1-y)}{1-i} \cdot \ds\frac{-3-4i}{(-3-x)+i(-2-y)}$

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$=\ds\frac{1-x+i(1-y)}{(-3-x)+i(-2-y)} \cdot \ds\frac{-3-4i}{1-i}
\cdot \ds\frac{1+i}{1+i}$

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$=\ds\frac{(1-x)+i(1-y)}{(-3-x)+i(-2-y)} \cdot \ds\frac{1-7i}{2}
\cdot \ds\frac{(-3-x)-i(-2-y)}{(-3-x)-i(-2-y)}$

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$=\ds\frac{(1-7i)}{2}$

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$\cdot
\ds\frac{[(1-x)(-3-x)+(1-y)(-2-y)+i((1-y)(-3-x)-(1-x)(-2-y))]}{(-3-x)^2+(-2-y)^2}$

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\un{Imaginary part is}: (numerator only)

$=-7[(1-x)(-3-x)+(1-y)(-2-y)]+(1-y)(-3-x)+(1-x)(+2+y)$

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$=-7(-3+2x+x^2+-2+y+y^2)+(-3+3y-x+xy+2-2+xy-xy)$

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$=21-14x-7x^2+14-7y-7y^2-1+4y-3x$

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$=-7x^2-17x-7y^2-3y+34$

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$=-7(x^2+\frac{17}{7}x)-7(y^2+\frac{3}{7}y)+34$

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$=-7((x+\frac{17}{14})^2-(\frac{17}{14})^2)-7((y+\frac{3}{14})^2-(\frac{3}{14})^2)+34$

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This equation describes a circle in $\bf{C}$; any point on this
circle gives a real cross ratio.  A number is real if and only if
its complex part is 0:

So $s$ gives a real cross ratio if and only if:

$$\un{\left(x+\ds\frac{17}{14}\right)^2+\left(y+\ds\frac{3}{14}\right)^2=\ds\frac{1250}{14^2}.\
\ \ (s=x+iy)}$$
\end{document}
