Minimum value of $|z-w|$ where $z,w \in \mathbb C$ such that $|z|=11$, and $|w+4+3i|=5$? I was thinking about the problem:

What is the minimum value of $|z-w|$ where $z,w \in \mathbb C$ such that $|z|=11$, and $|w+4+3i|=5$?

My attempts: I notice that $|z-w| \geq |z|-|w|=11-|w|$. Also if i take $w=u+iv$ then $|w+4+3i|=5$ gives $|w|^2+8u+6v=0$. Now i can not proceed. Can someone point me in the right direction? Thanks in advance for your time.
Let $z=a+bi$ and $w=c+di$. 

Now, $|z|=11\implies \sqrt{a^2+b^2}=11\implies a^2+b^2=121$, so $z$ is constrained to this circle (shown in blue below).

On the other hand, $|w+4+3i|=5\implies \sqrt{(c+4)^2+(d+3)^2}=5\implies (c+4)^2+(d+3)^2=25$, so $w$ is constrained to this circle (in red below).

The task is to minimize $|z-w|=\sqrt{(a-c)^2+(b-d)^2}$ subject to the two constraints above. Geometrically, we are looking to find the shortest distance between any two points which lie on the blue and red circles, respectively. A little calculus reveals the minimum is $1$ and is obtained when $a=-44/5$, $b=-33/5$ and $c=-8$, $d=-6$, as shown below.
$ |z-w|&gt; =  |z|-|w|  $

$|w+4+3i|=&lt;|w|+|4+3i|$

$|w|&gt;= 0$ $  Then Min|w|=0$

$|z-w|&gt;=(11-0)$

$|z-w|&gt;=11$

$Min|z-w|=11$