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?

You are correct that there isn't a unique circle passing through $a$ and $\frac{1}{\bar{a}}$, but the center of such a circle has to be equidistant from those two points, so on the perpendicular bisector of the segment between them.  Such a point can be described by $$c=\frac{1}{2}\left(a+\frac{1}{\bar{a}}\right)+ki\left(a-\frac{1}{\bar{a}}\right)$$ for some $k\in\mathbb{R}$ (that's the midpoint of the segment plus a scalar multiple of a $\frac{\pi}{2}$-rotation of the vector along that segment).  This can be simplified to $$c=\frac{1}{2}\left(a(1+2ik)+\frac{1}{\bar{a}}(1-2ik)\right).$$
The radius of the circle is $$\begin{align}
r=|c-a|&=\left|\frac{1}{2}\left(a(1+2ik)+\frac{1}{\bar{a}}(1-2ik)\right)-a\right|
\\
&=\frac{1}{2}\left|a(-1+2ik)+\frac{1}{\bar{a}}(1-2ik)\right|
\\
&=\frac{1}{2}\left|\frac{1}{\bar{a}}\left(a\bar{a}(-1+2ik)+(1-2ik)\right)\right|
\\
&=\frac{1}{2}\left|\frac{1}{\bar{a}}(1-a\bar{a})(1-2ik)\right|
\\
&=\frac{1}{2}\left|\frac{1}{\bar{a}}\right|\cdot|1-a\bar{a}|\cdot|1-2ik|
\\
&=\frac{|1-a\bar{a}|}{2|a|}\sqrt{4k^2+1}
.\end{align}$$
So now, as you suggested, we can use the converse of the Pythagorean Theorem: if the sum of the squares of the radii of the unit circle and our new circle is equal to the square of the distance between their centers, then they meet at right angles.  Mathematica tells me that the algebra works out and it's true, but I can't quite get there by hand.  Below is what I've got so far.
$$\begin{align}
\text{sum of squares of radii}&=1^2+\left(\frac{|1-a\bar{a}|}{2|a|}\sqrt{4k^2+1}\right)^2
\\
&=1+\frac{(1-a\bar{a})^2}{4a\bar{a}}(4k^2+1)
\end{align}$$
$$\begin{align}
(\text{distance }&\text{between centers})^2=|c-0|^2
\\
&=\left|\frac{1}{2}\left(a(1+2ik)+\frac{1}{\bar{a}}(1-2ik)\right)\right|^2
\\
&=\frac{1}{4}\left|a(1+2ik)+\frac{1}{\bar{a}}(1-2ik)\right|^2
\\
&=\frac{1}{4|\bar{a}|^2}|a\bar{a}(1+2ik)+(1-2ik)|^2
\\
&=\frac{1}{4a\bar{a}}|a\bar{a}(1+2ik)+(1-2ik)|^2
\end{align}$$
