Solution for $|z+3-4i|\leq5$ modulus of a complex number

Question

Here's my solution : \begin{align} &|z+3-4i|\leqq 5\\ &|x+3+(y-4)i|\leqq 5\\ &\sqrt{(x+3)^2+(y-4)^2)}\leqq 5\\ &(x+3)^2+(y-4)^2)\leqq 25 \end{align} And is now drawing a circle with $x=-3, y=4$ and $r=5$.
My answer will be all numbers lying on or inside the circle, is this a good answer?

Answer 1

Short Answer: Yes, your answer is correct.

Long Answer: Your answer could be more formal ("good answer"), in the sense that you could reference the inequality as an (representation)

circular area in the complex plane, bounded by $r = 5$ and centred at $(x, y) = (-3, 4)$

rather than just

a circle with $x = −3, y = 4$ and $r = 5$

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Furthermore, one could clarify the following:

"all numbers lying on or inside the circle"

with

Given $z = x + yi$, then all possible $(x, y)$ are satisfied by/within $(x + 3)^2 + (y - 4)^2 \le 25$