Step 1: Understand the equation.
We are given $\left|\frac{z-(2+i)}{z+(2-i)}\right|=2$. The modulus of a fraction is the modulus of the top divided by the modulus of the bottom. So this compares two distances.
Step 2: Split the modulus.
This becomes $|z-(2+i)|=2\,|z+(2-i)|$. The left side is the distance of $z$ from the point $(2,1)$. The right side is twice the distance of $z$ from $(-2,1)$.
Step 3: Put $z=x+iy$.
Then $|z-(2+i)|^2=(x-2)^2+(y-1)^2$ and $|z+(2-i)|^2=(x+2)^2+(y-1)^2$. We square both sides to remove the roots.
Step 4: Square the equation.
$(x-2)^2+(y-1)^2=4\left[(x+2)^2+(y-1)^2\right]$.
Step 5: Expand and gather.
Bring everything to one side. The squared terms $x^2$ and $y^2$ get the same factor $(1-4)=-3$ on each, so after subtracting they survive together with equal coefficients and no $xy$ term.
Step 6: Recognize the shape.
When the $x^2$ and $y^2$ terms survive with equal coefficients and there is no $xy$ term, the equation describes a circle. This is the well-known Apollonius circle that appears whenever the distance ratio is not $1$.
Step 7: Conclude.
Since the ratio here is $2$ (not $1$), the locus is a circle.
\[ \boxed{\text{A circle}} \]