Why do all circles passing through $a$ and $1/\bar{a}$ meet $|z|=1$ are right angles? In the complex plane, I write the equation for a circle centered at $z$ by $|z-x|=r$, so $(z-x)(\bar{z}-\bar{x})=r^2$. I suppose that both $a$ and $1/\bar{a}$ lie on this circle, so I get the equation
$$
(z-a)(\bar{z}-\bar{a})=(z-1/\bar{a})(\bar{z}-1/a).
$$
My idea to show that the circles intersect at right angles is to show that the radii at the point of intersection are at right angles, which is the case when the sum of the squares of the lengths of the radii of the circles is the square of the distance to the center of the circle passing through $a$ and $1/\bar{a}$. However, I'm having trouble finding a workable situation, since I don't think there is not a unique circle passing through $a$ and $1/\bar{a}$ to give a center to work with. What's the right way to do this?

Without lost of generality, we might suppose that $a$ is real, because we can roteate the plane as we want. By this suppose, $\frac{1}{\overline{a}}$ is also real since it is equal to $\dfrac{a}{|a|}$. 
It was already said that there are many circles passing for $a$ and $\frac{1}{\overline{a}}$ then we only can determine generically its centre $C$; $C=\frac{a+\frac{1}{a}}{2}+it$ for any $t\in \mathbb R$. Now we can write the equation $r^2=|z-\frac{a+\frac{1}{a}}{2}-it|^2(1)$. 
Let $w=\frac{a+\frac{1}{a}}{2},z=a$ and notincing that $\frac{1}{\overline{a}}=\frac{1}{a}$; we get from (1):
$$r^2=(a-w-it)(a-w+it)$$ $$=a^2-aw+ait-wa+w^2-wit-ait+iwt+t^2$$ $$=a^2-aw-wa+w^2+t^2$$ $$r^2=(a-w)^2+t^2.\quad(2)$$
As $|a|$ or $\frac{1}{a}|$ is less than 1, the circle $C_1:r^2=|z-\frac{a+\frac{1}{a}}{2}-it|^2$ intersects the unit circle $C_2:|z|=1$, so we can assume that there exists a $z\in C_1\cap C_2 \subset \mathbb C.$
Then for (1) and (2): $$(a-w)^2+t^2=(z-w-it)(\overline{z}-w+it)$$ $$(a-w)^2+t^2=|z|^2-zw+izt-\overline{z}w+w^2-iwt-i\overline{z}t+iwt+t^2$$ $$(a-w)^2=|z|^2-zw-\overline{zw}+w^2+izt+\overline{izt}$$ $$(a-w)^2=2|z|^2-zw-\overline{zw}+izt+\overline{izt}+w^2-|z|^2$$ $$(a-w)^2-w^2+|z|^2=2Re(|z|^2-zw +izt).$$
For the left side,
$$(a-w)^2-w^2+|z|^2=\frac{a^2-2+(\frac{1}{a})^2}{4}-\frac{a^2+2+(\frac{1}{a})^2}{4}+1=0$$
$\Rightarrow$ $$Re(|z|^2-zw +izt)=0$$
Remebering that the normal vector to the unit circle at the point $z$ is exactely $z$ and for $C_1$ at $z$ it is $z-\frac{a+\frac{1}{a}}{2}-it$. Let, $z,w\in \mathbb C$ the inner product in complex sense is $Re(z.\overline{w})$. Finally, $$Re((z-\frac{a+\frac{1}{a}}{2}-it ).(\overline{z}))=Re(|z|^2-zw -i\overline{z}t)=Re(|z|^2-zw +izt)=0.$$ Showing that the normal vector are making a right angle and so the tangents. $\Box$
