if $|z-3-2i| \leq 2 , \text{ find the minimum of } |2z-6+5i|$

Question

if $|z-3-2i| \leq 2 , \text{ find the minimum of } |2z-6+5i|$

My attempt:-

Let $z=x+iy$ so we have $(x-3)^2+(y-2)^2 \leq4$

let $|2z-6+5i | \text{ be }\phi $

so ${\phi}^2 =4(x-3)^2+(2y+5)^2$

I squared it as the first term is common, however,I'm not sure what to do from here

Working with @Martin R's suggestion, The region of The first number is a circle given by the cartesian equation $ (x-3)^2+(y-2)^2 \leq 4$

the point $(3,-2.5)$ lies 2.5 units below the circle, however, my books says the answer is 5,

Answer 1

The (reverse) triangle inequality helps: For all $z$ with $|z-3-2i| \leq 2$ is $$ |2z-6+5i| = 2 \left|z-3+ \frac 52 i\right| = 2 \left| z-3-2i + \frac 92 i \right| \\ \ge 2 \left( \left| \frac 92 i\right| - \left| z-3-2i\right|\right) \\ \ge 2 \left( \frac 92 - 2\right) = 5 \, . $$ Equality holds for $z=3$.

Geometrically, this is twice (!) the distance from the point $3-\frac 52i$ to the disk centered at $z=3+2i$ with radius $2$.