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.

Write $z=x+iy$, so $x^2+y^2=z\bar{z}$, and $x=\frac{z+\bar{z}}{2}$. Thus the circles can be described by
$$
z\bar{z}=a\frac{z+\bar{z}}{2}
$$
Upon doing $z=1/w$, you get
$$
\frac{1}{w\bar{w}}=\frac{a}{2}\frac{\bar{w}+w}{w\bar{w}}
$$
that becomes
$$
a\frac{\bar{w}+w}{2}=1
$$
Writing $w=X+iY$, you get
$$
aX=1
$$

More generally, if you have the circle $x^2+y^2+ax+by+c=0$, with $z=x+iy$ you get
$$
z\bar{z}+a\frac{z+\bar{z}}{2}+b\frac{z-\bar{z}}{2i}+c=0
$$
Upon changing $z=1/w$, the equation becomes
$$
\frac{1}{w\bar{w}}+a\frac{\bar{w}+w}{2w\bar{w}}
+b\frac{\bar{w}-w}{2iw\bar{w}}+c=0
$$
and removing $w\bar{w}$ from the denominator,
$$
1+a\frac{w+\bar{w}}{2}-b\frac{w-\bar{w}}{2i}+cw\bar{w}=0
$$
or, with $w=X+iY$,
$$
1+aX-bY+c(X^2+Y^2)=0
$$
So the circle becomes a straight line if $c=0$ (that is, it passes through the origin), otherwise it is transformed into a circle.
Similarly, the line $ax+by+c=0$ becomes
$$
a\frac{\bar{w}+w}{2w\bar{w}}+b\frac{\bar{w}-w}{2iw\bar{w}}+c=0
$$
and finally
$$
aX-bY+c(X^2+Y^2)=0
$$
so it becomes a circle if $c\ne0$, or a straight line through the origin otherwise.
