how to find greatest and least values of $|z-2-3i|$ if $|z-5-7i|=9$ Context: conceptual question in jee
I got the answer by using properties of modulus but i don't know if the procedure is correct

If $|z-5-7i|=9$ then find the greatest and least values of $|z-2-3i|$
 Answer: $14$, $4$

here's what I did: 

given,
$$||z|- |-5-7i||\leq |z-5-7i| \leq |z| + |-5-7i|$$
$$||z|- \sqrt{74}|\leq 9 \leq |z| + \sqrt{74}$$
idk if what i'm about to do next is correct I did this
$$9\leq|z|+\sqrt{74}=> |z| \geq 9-\sqrt{74}$$
I assumed $|z| = 9-\sqrt{74}$
$$||z|-|-2-3i|| \leq|z-(2+3i)|\leq|z|+|-2-3i|$$
$$|9-\sqrt{74}-5| \leq |z-(2+3i) \leq 9+5$$
$$|-4.60| \leq |z-(2+3i) \leq 14$$
Again I assumed $|-4.60|$ approx $4$
$$4\leq|z-(2+3i) \leq 14$$
Lookig at the question $|z-(5+7i)| = 9$ resembles equation of circle whose center at $5+7i$with radius 9, is there any way to solve this question using circle equation? and how does equation of cirlce co-relate to finding greatest and least value of $|z-2-3i|$

Is what i did correct above cause I got that answer by randomly by plugging value of $|z|$ as I can't find solid solution to this question in any module/internet/book.
 A: The set $c=\{z\in\Bbb C\mid|z-5-7i|=9\}$ is the circle centered at $C=5+7i$ with radius $9$. The points at which a point of that circle is nearest or further away from $P=2+3i$ are the points at which the line defined by $P$ and $C$ intersects the circle $c$; see the picture below. So, consider the points of the form $tC+(1-t)P$. Then\begin{align}\bigl|tC+(1-t)P-C\bigr|=9&\iff|(3+4i)(t-1)|=9\\&\iff|t-1|=\frac95\\&\iff t=\pm\frac{14}5\text{ or }t=-\frac45.\end{align}Can you take it from here?

A: Hint
I have to admit that I don't really understand what you did...
However, as you noticed the question is about finding the minimum and maximum distance between the point $p=2+3i$ and a point $z$ lying on the circle $\mathcal C$ of center $c=5+7i$ with a radius equal to $9$.
And the minimum and maximum distances are attained on the 2 points of intersection between the circle $\mathcal C$ and the line $\mathcal L$ passing through $p$ and $c$. A point of $\mathcal L$ is a complex $Z= (5+7i) + t [(5+7i)-(2+3i)]$ where $t \in \mathbb R$.
Plugging in $Z$ in the equation of the circle, you should find two values for  $t$ corresponding on the line to the extremas for the distance.
A: You could also treat this as an extremization problem in which you want the minimum and maximum values of $ \ (x - 2)^2 \ + \ (y - 3)^2  \ $ subject to the constraint $ \ (x - 5)^2  \ + \ (y - 7)^2 \ = \ 9^2 \ \ . $  The method of Lagrange multipliers (for example) gives us the equations
$$ 2·( x \ - \ 2) \ - \ \lambda \ · \ 2·(x \ - \ 5) \ \ = \ \ 0 \ \ , \ \ 2·( y \ - \ 3) \ - \ \lambda \ · \ 2·(y \ - \ 7) \ \ = \ \ 0  $$
$$ \lambda \ \ = \ \ \frac{x \ - \ 2}{x \ - \ 5} \ \ = \ \ \frac{y \ - \ 3}{y \ - 7} \ \ \Rightarrow \ \ xy \ - \ 7x \ - \ 2y \ + \ 14 \ \ = \ \ xy \ - \ 3x \ - \ 5y \ + \ 15  $$
$$ \Rightarrow \ \ 4x \ - \ 3y \ \ = \ \ -1 \ \ , \ \ x \neq 5 \ , \ y \neq \ 7$$
(this is the line depicted in José Carlos Santos's graph).  Inserting this into the equation for the constraint circle yields
$$ 16·(x - 5)^2  \ = \ 16·81 \ - \  16·(y - 7)^2 \ \ \rightarrow \ \ (4x - 20)^2  \ = \ 16·81 \ - \  16·(y - 7)^2  $$
$$ \Rightarrow \ \ (3y - 21)^2  \ = \ 16·81 \ - \  16·(y - 7)^2 \ \ \Rightarrow \ \ 9·( y - 7)^2  \ = \ 16·81 \ - \  16·(y - 7)^2  $$
$$ \Rightarrow \ \ 25·( y - 7)^2  \ = \ 16·81 \ \ \Rightarrow \  \ y \ \ = \ \ 7 \ \pm \ \sqrt{\frac{16·81}{25}} \ \ = \ \ 7 \ \pm \ \frac{36}{5}   $$
$$ \Rightarrow \ \ x \ \ = \ \ \frac{3y \ - \ 1}{4} \ \ = \ \ \frac{3·\left(7 \ \pm \ \frac{36}{5} \right) \ - \ 1}{4} \ \ = \ \ \frac{20  \ \pm \ \frac{108}{5}  }{4} \ \ = \ \ 5 \ \pm \ \frac{27}{5} \ \ . $$
We now have the coordinates of the two points of tangency for the circle centered on $ \ 2 + 3i \ $ with the circle centered on $ \ 5 + 7i \ $ when the former circle has its maximum and minimum radii, respectively.  Using these coordinates in the "radius-squared" function gives us
$$ \left(5 \ \pm \ \frac{27}{5} \ - \ 2 \right)^2 \ + \ \left(7 \ \pm \ \frac{36}{5} \ - \ 3 \right)^2 \ \ = \ \  \left(3 \ \pm \ \frac{27}{5}  \right)^2 \ + \ \left(4 \ \pm \ \frac{36}{5} \right)^2 $$ $$ = \ \ (3^2 \ + \ 4^2) · \left(1 \ \pm \ \frac{9}{5}  \right)^2  \ \ = \ \  5^2 · \left(1 \ \pm \ \frac{9}{5}  \right)^2 \ \ . $$
Hence, the maximum and minimum values we've sought are $ \ 5 ·  \frac{14}{5}  = \ 14 \  \ $ and $ \ 5 ·  \frac{4}{5}  = \ 4 \  \ . $
