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?

I have a solution that relies on converting the complex numbers into ordered pairs although I believe there must be a solution with just the help of complex numbers.
Two circles intersect orthogonally, if their radii are perpendicular at the point of intersection. So, using this we can have a condition for orthogonality. 
$\hskip 2.5in$ 
Here's a trick of how you will get a condition. Let us consider two circles,
$$C_1,A:x^2+y^2+2g_1x+2f_1y+c_1=0$$
$$C_2,B:x^2+y^2+2g_2x+2f_2y+c_2=0$$
From your high school course in analytical geometry in high school, it must be clear that the centres $A$ and $B$ are $A(-g_1,-f_1)$ and $B(-g_2,-f_2)$. And the radii of such a circle is $r_1=\sqrt{g_1^2+f_1^2-c_1}$ and similarly $r_2=\sqrt{g_2^2+f_2^2-c_2}$. 
Now invoke Pythagoras here, I'll leave the actual computation to you, the condition would turn out to be, $$2g_1g_2+2f_1f_2=c_1+c_2$$
Now, find a parametric equation for a circle passing through the complex numbers $a$ and $\dfrac{1}{\bar a}$. How do you do this? 
Since the circle always passes through, $a\cong(l,m)$ and $\dfrac{1}{\bar a}=\dfrac{a}{|a|^2}\cong\left(\dfrac{l}{l^2+m^2},\dfrac{m}{l^2+m^2}\right)$, you have the following will be the equation of the circle:
$$(x-l)\left(x-\dfrac{l}{l^2+m^2}\right)+(y-m)\left(y-\dfrac{m}{l^2+m^2}\right)+\lambda(ly-mx)=0$$
The second circle is, $$x^2+y^2-1=0$$
So, you should now see that $g_2=f_2=0$ and $c_2=-1$. Also, after a little inspection, note that we need not care for what those $g_1$ and $f_1$ are. And, thankfully, $c_1=1$. So, you have the required condition for orthogonality.
I know this is lengthy and not instructive, but this is all I can recollect from high school geometry. So, I only hope this is of some help!
