# The image of a circle under $z \mapsto \frac{z- \frac{1}{2}}{\frac{1}{2}z-1}$

Let $C_1 = \{ z: |z| = \frac{1}{7} \}$ be a circle inside the unit circle $C_0 = \{ z: |z|=1\}$. The fractional linear transformation: $$\phi: z \mapsto \frac{z- \frac{1}{2}}{\frac{1}{2}z-1}$$ is a map from the unit cirle to itself, $\phi: C_0 \to C_0$. What is the image of $C_1$ ? Obviously it is a circle, but what is the radius and center of $\phi(C_1)$?

I hope this figure not too confusing. $\phi$ maps the light blue circle to the dark blue circle. Similarly it maps light green to dark green.

In regard to 3 points... I know there is formula for 3 points $(x_1, y_1), (x_2, y_2),(x_3, y_3)$ going through a circle using a determinant but I have not used it:

$$\left|\begin{array}{cccc} x^2 + y^2 & x & y & 1 \\ x_1^2 + y_1^2 & x_1 & y_1 & 1 \\ x_2^2 + y_2^2 & x_2 & y_2 & 1 \\ x_3^2 + y_3^2 & x_3 & y_3 & 1 \\ \end{array} \right|=0$$

Which 3 points should I pick anyway? Even more alternatively, are there ways to solve this using power of a point from geometry?

I suspect @vvnitram's answer is wrong. Here is a plot of the image circle (computed numerically) and the circle he proposes. These are not quite the same.

• Think of three points on $C_1$. Find their images under $\phi$. See what circle goes through those points. – GEdgar Aug 3 '16 at 0:21
• Why don't you find three image points, and then find the unique circle that passes through these points? – Dimitris Aug 3 '16 at 0:21
• See where $\phi$ maps each of the following three points: $\pm1/7$, $i/7$. – avs Aug 3 '16 at 0:22

Let $$w=u+iv=\frac{2z-1}{z-2}$$

Rearranging, we get $$z=\frac{2w-1}{w-2}$$

So the image of the circle $|z|=\frac 17$ is given by $$\left|\frac{2w-1}{w-2}\right|=\frac 17$$ $$\Rightarrow 7|2w-1|=|w-2|$$ $$\Rightarrow 49((2u-1)^2+4v^2)=(u-2)^2+v^2$$

This simplifies to $$195u^2+195v^2-192u+45=0$$

This is a circle, centre $(\frac{32}{65},0)$ and radius $\frac {7}{65}$

Here is a quicker approach. It is clear that in this Mobius transformation a circle is mapped onto a circle (not through Origins). So a diameter of the given circle is mapped onto a diameter of the image circle. Take two diametrical points: $(1/7,0)$ and $(-1/7,0)$ and subject those to the given transformation. Little arithmetic shows $(5/13,0)$ and $(3/5,0)$. The distance between these points is easy to see: $14/65$ which is the diameter of the image circle

• That's the best solution (ofc i would say that, it's my solution too). +1 – zhw. Aug 4 '16 at 17:58
• @zhw. Thanks, that's the "trick" we learned when I took this class...and from here the equation of David Quinn can also be easily derived... – imranfat Aug 4 '16 at 17:59
• Right. Letting $c=(5/13 + 3/5)/2, r= (3/5-5/13)/2,$ the eqn is simply $|z-c|= r.$ – zhw. Aug 4 '16 at 18:02

Since at this point I am solving my own question, let me note that:

$$\phi: \frac{1}{7} \mapsto \frac{ \frac{1}{7}- \frac{1}{2}}{\frac{1}{2}\times\frac{1}{7}-1}, \hspace{0.25in} -\frac{1}{7} \mapsto \frac{ -\frac{1}{7}- \frac{1}{2}}{-\frac{1}{2}\times\frac{1}{7}-1}, \hspace{0.25in} \frac{i}{7} \mapsto \frac{ \frac{i}{7}- \frac{1}{2}}{\frac{1}{2}\times\frac{i}{7}-1}$$

Those are fine and then I have to solve the determinant equation:

$$\left|\begin{array}{cccc} x^2 + y^2 & x & y & 1 \\ x_1^2 + y_1^2 & x_1 & y_1 & 1 \\ x_2^2 + y_2^2 & x_2 & y_2 & 1 \\ x_3^2 + y_3^2 & x_3 & y_3 & 1 \\ \end{array} \right|=0$$

Here's re-write in terms complex numbers $z = x+iy$ and $z = x - iy$. I did not to the change of basis, I just am guessing:

$$\left|\begin{array}{rlll} |z|^2 & z & \overline{z} & 1 \\ |z_1|^2 & z_1 & \overline{z}_1 & 1 \\ |z_2|^2 & z_2 & \overline{z}_2 & 1 \\ |z_3|^2 & z_3 & \overline{z}_3 & 1 \\ \end{array} \right|=0$$

In my case, $z_1 = \phi(\frac{1}{7})$ and $z_1 = \phi(-\frac{1}{7})$ and $z_1 = \phi(\frac{i}{7})$ .

Note that $\phi(0)=1/2$ and $\phi(1/7)=5/13$.

Because is a moebious transform, send the center into center. Then, $\phi(C_1)=C$ where $C$ is the circunference with center $1/2$ and radious $1/2-5/13=3/26$

• Mobius transformations do NOT send centers of circles to centers. – Dimitris Aug 3 '16 at 0:22
• Yes, in this case. Because sends simmetric points into simmetric points, and the simmetric respect the center is \infty – Martín Vacas Vignolo Aug 3 '16 at 0:23
• No, this map does not send the center to the center. – zhw. Aug 4 '16 at 17:59