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.

$$|z-3|=5.$$
Let $\displaystyle w= \frac{1}{z}$ and let $w^\star$ be the conjugate of $w$.
$$\begin{aligned} 
\left(\frac{1}{w} -3 \right) \left(\frac{1}{w^\star} -3 \right) &= 25 \\
(1 -3 w^\star) \left( 1 - 3w \right) &= 25 w w^\star\\
1-3(w+w^\star) + 9 w w^\star &= 25 w w^\star \\
16 w w^\star + 3(w + w^\star) -1 &= 0\\
w w^\star + \frac{3}{16} (w + w^\star) &= \frac{1}{16}\\
\left( w + \frac{3}{16}\right) \left( w^\star + \frac{3}{16} \right) &= \frac{9}{16^2} + \frac{1}{16} = \left( \frac{5}{16}\right)^2\\
\left| w + \frac{3}{16} \right| &= \frac{5}{16}.
\end{aligned}$$
