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 $z=11(\cos A+i\sin A), w=r(\cos B+i\sin B)$ so $|w+4+3i|=\sqrt{(r\cos B+4)^2+(r\sin B+3)^2}=\sqrt{r^2+25+2r(4\cos B+3\sin B)}$
So, $r^2+25+2r(4\cos B+3\sin B)=25\implies r=-2(4\cos B+3\sin B)$ as $r\ne0$
$|z-w|^2=(11\cos A-r\cos B)^2+(11\sin A-r\sin B)^2=11^2+r^2-22r\cos(A-B)\ge (11-r)^2$
Now, let $R\sin C=4,R\cos C=3$ where $R>0$ so that $R^2=4^2+3^2=5^2\implies R=5$ and $\sin C=\frac45,\cos C=\frac35$
so $4\cos B+3\sin B=R\sin(B+C)=5\sin(B+C)$
So, $-5\le 4\cos B+3\sin B\le 5\implies -10\le r\le 10$
$\implies|z-w|^2\ge (11-10)^2=1$ which occurs if $A=B$ and if $B+C=-\frac\pi2\implies \sin B=\sin(-\frac\pi2-C)=-\sin(\frac\pi2+C)=-\cos C=-\frac34$ and $\cos B=\cos(-\frac\pi2-C)=\cos(\frac\pi2+C)=-\sin C=-\frac45$
So, $z=-\frac{11}5(4+3i),w=-\frac{10}5(4+3i)=-(8+6i)$
