Show that by means of the transformation $w=\frac 1z$ the circle $C$ given by $|z-3|=5$ is mapped into the circle $|w+\frac{3}{16}|=\frac{5}{16}$ 
Show that by means of the transformation $w=\frac 1z$ the circle C given by $|z-3|=5$ is mapped into the circle $\left|w+\frac{3}{16}\right|=\frac{5}{16}$

My try:
$$\begin{align}\\
&w=\frac 1z\\
&\implies 3w=\frac 3z\\
&\implies1-3w=1-\frac 3z=\frac{z-3}{z}\\
&\implies|1-3w|=\left|\frac{z-3}{z}\right|\\
&\implies|1-3w|=|z-3|\left|\frac 1z\right|\\
&\implies|1-3w|=5|w|\\
\end{align}\\$$
But, I can't understand what to do next and how can I proof that. Any suggestions? or, you can add an answer.

We need to prove that if $|z-3|=5$ so $$\left|\frac{1}{z}+\frac{3}{16}\right|=\frac{5}{16}.$$
Let $z=x+yi,$ where $x$ and $y$ are reals.
Thus, $$(x-3)^2+y^2=25$$ and we need to prove that:
$$|3z+16|=5|z|$$ or
$$(3x+16)^2+9y^2=25(x^2+y^2)$$ or
$$16(x^2+y^2)=96x+256$$ or
$$x^2-6x+y^2=16$$ or
$$(x-3)^2+y^2=25$$ and we are done!
