Every line or circle in $\mathbb{C}$ are solution sets to the equation... Here is a complex analysis homework problem I can't quite figure out:

Prove that every line or circle in $\mathbb{C}$ is the solution set of an equation of the form $a|z|^2+\bar{w}z+w\bar{z}+b=0$, where $a,b\in\mathbb{R}$ and $w,z\in\mathbb{C}$. Conversely, show that every equation of this form has a line, circle, point, or the empty set as its solution set.

So far, I've tried to rewrite the equation of a line in $\mathbb{R}^2$ as $y=mx+b$ in $\mathbb{C}$, where $m$ is real and $x,b$ are complex. I know that a circle in the complex plane is given by $|z-a|=r$, where $a$ is the center and $r$ is the radius.
I also noticed that $\bar{w}z+w\bar{z}=2\text{Re}(\bar{w}z)$. I'm just not sure how all these pieces fit together in answering the question. Any help would be greatly appreciated.
 A: You need to use the property of complex conjugation to express the circle equation 
\begin{align}
|z - z_0|^2 &= r^2 \Leftrightarrow  \overline{(z - z_0)}(z-z_0) = r^2 \Leftrightarrow (\overline{z} - \overline{z_0})(z - z_0) = r^2 \Leftrightarrow \\
& z\overline{z} - \overline{z}z_0 - z\overline{z_0} + |z_0|^2 = r^2
\end{align}
Now make the substitution $z_0 = -\frac{\overline{\alpha}}{A}$ where $A \in \mathbb{R} \setminus \{0\}$
$$
z\overline{z} + \overline{z}\frac{\overline{\alpha}}{A} + z\frac{\alpha}{A} + \left|\frac{\overline{\alpha}}{A}\right|^2 - r^2 = 0 
$$
Mutliplying the equation by $A$
$$
Az\overline{z} + \overline{z\alpha} + z\alpha + A\left(\left|\frac{\alpha}{A}\right|^2 - r^2\right) = 0 
$$
And setting $B = A\left(\left|\frac{\alpha}{A}\right|^2 - r^2\right)$ yields the general form of the equation for a circle in the complex plane. This equation also describes lines which can be viewed as circles with infinite radius. 
$$
Az\overline{z} + \overline{z\alpha} + z\alpha + B = 0 
$$
When $A = 0$ it represents a line, 
$$
\overline{z\alpha} + z\alpha + B = 0 
$$
You can convice yourself that this equation describes a line by setting $z = x + iy$ and $\alpha = p + iq$.  
A: A line or circle in the plane has an equation of the form
$$
D(x^2+y^2)+Ax+By+C=0.
$$
It's a line if $D=0$, but $A$ and $B$ are not both zero; it's a circle if $D\ne0$ and $A^2+B^2-4CD^2>0$. So the equation above represents either a line or a circle if $A^2+B^2-4CD^2>0$. The equation represents a single point if $A^2+B^2-4CD^2=0$ but $A$ and $B$ are not both zero (so $D\ne0$). It represents the empty set when $A^2+B^2-4CD^2<0$. (I'll assume that not all coefficients are zero.)
Set $z=x+iy$; then
$$
x=\frac{z+\bar{z}}{2},\quad
y=\frac{z-\bar{z}}{2i}=-i\frac{z-\bar{z}}{2},\quad
x^2+y^2=z\bar{z}.
$$
Substituting in the equation above we get
$$
2Dz\bar{z}+A(z+\bar{z}-iB(z-\bar{z})+2C=0
$$
or
$$
2Dz\bar{z}+(A-iB)z+(A+iB)\bar{z}+2C=0.
$$
Setting $2D=a$, $A+iB=w$ and $2C=b$ we get the requested form.
The converse is just doing the converse substitution: set $D=a/2$, $C=b/2$, $A=(w+\bar{w})/2$ and $B=(w-\bar{w})/(2i)$. Also $A^2+B^2=w\bar{w}$.
The condition for a line/circle reads
$$
w\bar{w}-4\frac{b}{2}\frac{a^2}{4}>0
$$
or
$$
2w\bar{w}-a^2b>0.
$$
Single point when $2w\bar{w}=a^2b\ne0$, empty set when $2w\bar{w}-a^2b<0$.
A: The equations for circles and lines over $\mathbb{R}$ are quadratic equations where the quadratic term is of the form $ax^2+ay^2$. From your observations about the reality of the linear term the claim follows.
