Prove that the map $f(z)=\frac{1}{z}$ sends any line onto either a line or a circle. Show the cases in which the image is a line and the case in which the image is a circle. 
I understand that representing the equation of line, ($ax+by+c=0$  $a,b,c\in\Bbb R$  $a,b\neq0$ at the same time), as $pz+\overline {pz} +2c=0$ where $p$  is a suitable nonzero complex number, should help me achieve my end goal in this proof. However, I'm not sure how to use this fact to do so as well as how to justify why I should think of the line in this way.

First of all, I think that the rquation of the line $ax + by + c = 0$, when written in terms of $z = x + iy$, should be
$\bar p z + p \bar z + 2c = 0, \tag{1}$
and not
$pz + \bar p z + 2c = 0; \tag{2}$
here $p = a + bi$.  To see the problem with (2), note that it may be written
$(p + \bar p)z + 2c= 0, \tag{3}$
and that
$p + \bar p = (a + bi) + (a - bi) = 2a; \tag{4}$
thus (2) becomes
$2az + 2c = 0. \tag{5}$
Assuming $a \ne 0$, the only solution of (5) is
$z = -\dfrac{c}{a}; \tag{6}$
we see that in this case (5) represents the single point (6), not the line
$ax + by + c = 0, \tag{7}$
and if $a = 0$, then (5) forces $c = 0$ and any $z \in \Bbb C$ solves (5); in neither case does (2) represent a line.  (1), however, is easily seen to be equivalent to (7), as follows:  with $p$ and $z$ as above, it follows that
$\bar p z = (a - bi)(x + iy) = (ax + by) + i(ay - bz), \tag{8}$
and hence, since
$p \bar z = \overline{\bar p z} \tag{9}$
we have from (8) that
$\bar p z + p \bar z = 2(ax + by), \tag{10}$
whence
$\bar p z + p \bar z + 2c = 2ax + 2by + 2c; \tag{11}$
it is easy to see from (11) that (1) and (7) represent the same line. 
Granting that the correct equation for a line, in terms of $z = x + iy$, is (1), we investigate the action of $f(z) = 1/z$.  If $c \ne 0$, then clearly (1) does not pass through the origin; $z \ne 0$ for all $z$ satisfying (1).  Then $z \bar z \ne 0$ as well, and we may write
$f(z) = \dfrac{1}{z} = \dfrac{\bar z}{z \bar z} \tag{12}$
for any such $z \in \Bbb C$.  If we apply $f(z)$ in the form of (12) to (1) we obtain
$\bar p \dfrac{1}{z} + p \dfrac{1}{\bar z} + 2c = 0 \tag{13}$
or
$\bar p \dfrac{\bar z}{z \bar z} + p \dfrac{z}{z \bar z} + 2c = 0; \tag{14}$
multiplying (14) though by $z \bar z = x^2 + y^2 \ne 0$ yields
$\bar p \bar z + p z + 2c z \bar z = 0, \tag{15}$
which I claim is the equation of a circle provided $c \ne 0$.  To wit:  with $z = x + iy$ and $p = a + bi$,
$pz = (a + bi)(x + iy) = (ax - by) + i(ay + bx), \tag{16}$
and since
$\bar p \bar z = \overline{pz} \tag{17}$
we see that
$\bar p \bar z + p z + 2c z \bar z = 2(ax - by) + 2c(x^2 + y^2), \tag{18}$
and with $c \ne 0$ we may thus infer
$\dfrac{a}{c}x - \dfrac{b}{c}y + x^2 + y^2 = 0; \tag{19}$
we complete the squares in $x$ and $y$ separately by adding $(a^2 + b^2)/4c^2$ to both sides:
$(x + \dfrac{a}{2c})^2 + (y - \dfrac{b}{2c})^2 = \dfrac{a^2 + b^2}{4c^2}. \tag{20}$
Since by hypothesis $a^2 + b^2 \ne 0 \ne c$, (20) describes a circle of non-zero radius $\sqrt{(a^2 + b^2/4c^2}$ centered at $(-a/2c, b/c)$.  This covers the case $c \ne 0$.
If $c = 0$, then $z = 0$ is a solution to (1) and we may only use (12) away from $0$; indeed, caution must be taken with the transformation $f(z) = 1/z$ since $0$ must be handled separately.  For $z \ne 0$ we may set $c = 0$ in (18) and thus deduce that
$ax - by = 0; \tag{21}$
this is the equation of a line passing through the origin, normal to the vector $(a, - b)$, since (21) may be written
$(a, - b) \begin{pmatrix} x \\ y \end{pmatrix} = 0. \tag{22}$
Apparently, away from $0$, the transformation $1/z$ maps the line $ax + by = 0$ (from (1), (10)) to the line $ax - by = 0$; $0$ maps to the so-called "point at infinity", which maps back to $0$.  We summarize:
$f(z) = \dfrac{1}{z}$ maps
Lines not passing through $0$ to circles;
and 
Lines passing throug $0$ to other lines passing through $0$.
