Show that all circles that pass through a and 1/ā intersect the circle |z|=1 at right angles
In the complex plane, show that all circles that pass through \(a\) and \(1/\bar{a}\) intersect the circle \(\lvert z\rvert = 1\) at right angles.
Answer
Since \(1/\bar{a}=a/|a|^2\) has the same argument as \(a\), the line through \(a\) and \(1/\bar{a}\) (the dotted line in the above figure) passes through the origin.
Let the centre of the circle which passes through \(a\) and \(1/\bar{a}\) be \(c\) and its radius \(r\), then the equation of the circle is \(|z-c|=r\), and \(|a-c|=r=|1/\bar{a}-c|\).
First use Pythagoras's theorem to find a relation between \(r\) and \(|c|\). The centre, \(c\), is equidistant from \(a\) and \(1/\bar{a}\), so it lies on the perpendicular bisector of the line from the origin to \(a\) which meets it at the midpoint between \(a\) and \(1/\bar{a}\), \(m\equiv\frac12(a+1/\bar{a})\).
Denote the distance from \(m\) to \(c\) by \(k\) then by Pythagoras, \(|c|^2=|m|^2+k^2\) and \(r^2=|m-1/\bar{a}|^2+k^2\), so eliminating \(k\), \(|c|^2=r^2+|m|^2-|m-1/\bar{a}|^2\). Then \( m-1/\bar{a}=\frac12(a-1/\bar{a}),\) and so \[ |m|^2-|m-1/\bar{a}|^2=a\bar{a}/|a|^2=1, \] so finally \(|c|^2=r^2+1\).
Let \(z\) be a point of intersection of the two circles, then the triangle formed by \(O,c\) and \(z\) has sides \(r\), \(1\), and \(|c|=\sqrt{1+r^2}\) and is thus by Pythagoras a right angled triangle, and so since the circumference of a circle is orthogonal to its radius, the two circles meet at right angles.
Another way to solve the problem
Another way of showing that the radii are perpendicular with less geometry and more algebra is to show that their divisor is purely imaginary. Let \(z\) be a point of intersection of the unit circle \(|z|=1\) and \(|z-c|=r,\) then
\[ r^2= |z-c|^2= |z|^2-(c\bar{z}+\bar{c}z)+|c|^2= 1-(c\bar{z}+\bar{c}z)+|c|^2= 2-(c\bar{z}+\bar{c}z)+r^2 \]which yields \(c\bar{z}+\bar{c}z=2\), so using \(|c\bar{z}|=|c||z|=|c|=\sqrt{1+r^2}\), \(c\bar{z}=1\pm ir\). The radius from \(c\) to \(z\) is \(z-c\), and the ratio of this radius to the radius of the inner circle, \((z-c)/z\) = \(c/z-1=c\bar{z}-1=\pm ir\) is purely imaginary, thus the lines are perpendicular.