Help with proving that a circle containing two complex numbers, $w=x+iy$ and $1/(\bar{w})$ must intersect ${|z|=1}$ at right angles? Question: Proving that a circle containing two complex numbers, $w=x+iy$ and $1/(\bar{w})$ must intersect the circle, ${|z|=1}$ at right angles?
I am currently in complex analysis and as a way to further our understanding of the topics he assigned us each different questions that would further our knowledge outside of what was taught in class. I am not looking for the actual proof as that would be counter productive. However, I am interested in things that could help me along in trying to even understand how this would be true. 
The only equation for a circle in the complex plain we have so far is $|z-z_0|=p$  where $z_0$ and $p$ are the center and the radius of the circle, respectively. I have the feeling that I will need more than this to develop an equation of a circle containing the two points mentioned above. However I am not sure if this is the correct way to even think about it. He always emphasizes the importance of geometric thinking. However, I am severely under developed in being able to think about these concepts in that light and prefer falling back to the algebra. 
Things I have found that may be useful so far:
1) The argument of both $w$ and $1/\bar{w}$ are the same.
2) Similar questions in $\mathbb{R}^2$ on this sight have referred to taking the determinants. However I am not sure of how much help this will be.
 A: This is probably ahead of your class, but material you are guaranteed to be shown. A Mobius transformation takes four constants $A,B,C,D$, real or complex as needed, and gives the mapping
$$ f(z) = \frac{Az + B}{Cz + D}  $$
There are several properties: a Mobius transformation maps a circle to either a line orr a circle, and it maps a line to either a line or a circle. Also, a Mobius transformation preserves angles.
The transformation 
$$  f(z) = \frac{z+i}{iz+1} $$
takes the circle $|z|=1$ to the real line. Worth checking, as you like algebra. The same transformation takes $w$ and $1/\bar{w}$ to conjugate points. If $f(w) = G$ and $f(\frac{1}{\bar{w}} = H,$ then
$$  H = \bar{G} $$
If you draw a little diagram, you will see that any circle through $G,H$ is orthogonal to the real line. Also, the straight line between $G,H$ is orthogonal to the real line. 
Let's see: Mobius transformations have easy inverses: in this case it is also the formal reciprocal,
$$  g(z) = \frac{iz+1}{z + i} \; . $$
Given some complex $t,$ what is the relationship between $g(t)$ and $g( \bar{t}) \; ?$ For that matter, what can you say about $g(t)$ when $t$ is real?
