I was trying to solve this problem ,
so $y = c$. Suppose for ease let us take $c > 0$.
so $w = \frac{1}{z} = \frac{1}{x + iy} = \frac{x - iy}{x^2 + y^2} = \frac{x - ic}{x^2 + c^2} = \frac{x}{x^2 + c^2} - i( \frac{c}{x^2 + c^2})$.
Now as $w = u + iv$,
comparing we get $u = \frac{x}{x^2 + c^2}$ and $v = \frac{-c}{x^2 + c^2}$ and also $u^2 + v^2 = \frac{1}{x^2 + c^2} \leq \frac{1}{c^2}$ and I think this is a disc centered at $0$ and of maximum radius of $\frac{1}{c}$.
Next I am thinking that since $v = \frac{-c}{x^2 + c^2}$ and $c > 0$ that would imply that $v$ is always negative but now if this sis the case then what I have got the region as disc will be wrong as it contains some positive $v$ also the upper part of the disc and hence now I am getting the region as semicircular disc with arc in the negative $v$ axis.
Is this correct?.if not where I am making mistake. Any help is great.
Hint: Consider the absolute value of $w+\frac i{2c}$.
Also look at Inversive Geometry.