Image of a family of circles under $w = 1/z$ Given the family of circles $x^{2}+y^{2} = ax$, where $a \in \mathbb{R}$, I need to find the image under the transformation $w = 1/z$. I was given the hint to rewrite the equation first in terms of $z$, $\overline{z}$, and then plug in $z = 1/w$.
However, I am having difficulty doing this. 
I completed the square in $x^{2}+y^{2}=ax$ to obtain $\left(x - \frac{a}{2} \right)^{2} + y^{2} = \left(\frac{a}{2} \right)^{2}$.
Then, given that $\displaystyle x = Re(z) = \frac{z+\overline{z}}{2}$ and $\displaystyle y = Im(z)= \frac{-(z-\overline{z})}{2i}$, I made those substitutaions and my equation became $\left( \frac{z+\overline{z}-a}{2}\right)^{2} - \left(\frac{\overline{z}-z}{2} \right)^{2} = \left(\frac{a}{2} \right)^{2}$.
Then, sbustituting in $z = \frac{1}{w}$, this became $\displaystyle \frac{\left(\frac{1}{w} + \frac{1}{\overline{w}} - a \right)^{2}}{4} - \frac{\left(\frac{1}{\overline{w}} - \frac{1}{w} \right)^{2}}{4} = \frac{a^{2}}{4}$.
Beyond this, my algebra gets very wonky.
Could someone please tell me what my final result should be? Knowing that would allow me to work backwards and then apply these methods to other problems (of which I have many to do!).
Thanks.

Hint: This transformation can be seen as successive transformations  $$Z = \frac{z}{|z|^2} , \,\,\,w = \overline {Z}$$ 
since $z \cdot \overline{z} = |z|^2$.
Then $z = \frac{1}{w} = \frac{\overline w}{|w|^2}$ and $$x = \frac{u}{u^2 + v^2} \,\,\,\text{and} \,\,\,\,y = \frac{-v}{u^2+v^2}$$
where $w = u + iv$.
Edit: Now study $$A(x^2+y^2) + Bx + Cy + D = 0$$
complete the square and make the substitutions to see that is taken to 
$$D(u^2 + v^2) + Bu - Cv + A = 0 $$
which is a circle or a line., depending on whether $D \neq  0$ or $D=0.$ 
See what happens when a cirle $A \neq 0$ is passing or not through the origin.
