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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?

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The line segment from $a$ to $1 / \bar{a}$ has a perpendicular bisector. The center of any circle being discussed is on that line. – Will Jagy Jan 26 '12 at 22:12

This is a basic property of Inversion of a point through a circle.

Suppose you are given a circle of radius $r$ with center $O$ (called the reference circle) and a point $P$ not on the circle, then the inverse of $P$ with respect to the reference circle is given by the point $P'$ which is collinear with $O$ and $P$ (on the ray $\overrightarrow{OP}$) such that $OP \times OP' = r^2$. Points on the circle get mapped to themselves.

A property of inversion is that a circle orthogonal to the reference circle is its own inverse and a circle which is its own inverse is orthogonal to the reference circle.

This follows easily from the Intersecting Secants Theorem (see the link for a nice applet), whose proof only uses the fact that the angle subtended by a chord on any point on the major(minor) arc of the circle is constant, and thus the triangles formed are similar. A proof can be found here. These proofs are very simple and do not require any cumbersome analytic geometry and is usually taught in high school geometry courses.

Now $a$ and $\frac{1}{\overline{a}}$ are inverse points with respect to the circle $|z| = 1$ and thus any circle passing through those two points is its own inverse (why?), and as a consequence, is orthogonal to the unit circle.

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Building on Isaac's answer, I think you can get the identity you're looking for by doing this:

\begin{eqnarray*} 1+|c-a|^2 &=& 1+(c-a)(\overline{c}-\overline{a})\\ &=& 1+|c|^2+|a|^2-2\operatorname{Re}(c\overline{a}) \\ &=& 1+|c|^2+|a|^2-2\operatorname{Re}\left[\overline{a}\left(\frac{1}{2}\left(a+\frac{1}{\overline{a}}\right)+ik\left(a-\frac{1}{\overline{a}}\right)\right)\right] \\ &=& 1+|c|^2+|a|^2-2\operatorname{Re}\left[\frac{1}{2}\left(|a|^2+1\right)+ik\left(|a|^2-1\right)\right]\\ &=& 1+|c|^2+|a|^2-2\left[\frac{1}{2}(|a|^2+1)\right] \\ &=& 1+|c|^2+|a|^2-|a|^2-1 \\ &=& |c|^2 \end{eqnarray*}

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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.

$\hskip 2.5in$ Orthogonal Circles

Here's a trick of how you will get a condition. Let us consider two circles,



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:


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!

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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. $$\begin{align} \text{sum of squares of radii}&=1^2+\left(\frac{|1-a\bar{a}|}{2|a|}\sqrt{4k^2+1}\right)^2 \\ &=1+\frac{(1-a\bar{a})^2}{4a\bar{a}}(4k^2+1) \end{align}$$

$$\begin{align} (\text{distance }&\text{between centers})^2=|c-0|^2 \\ &=\left|\frac{1}{2}\left(a(1+2ik)+\frac{1}{\bar{a}}(1-2ik)\right)\right|^2 \\ &=\frac{1}{4}\left|a(1+2ik)+\frac{1}{\bar{a}}(1-2ik)\right|^2 \\ &=\frac{1}{4|\bar{a}|^2}|a\bar{a}(1+2ik)+(1-2ik)|^2 \\ &=\frac{1}{4a\bar{a}}|a\bar{a}(1+2ik)+(1-2ik)|^2 \end{align}$$

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