1. Interpret the given equation geometrically

Given $2|z+3i|-|z-i|=0$ we get $2|z+3i|=|z-i|$

Let $z=x+iy$ Then $|z+3i|=|x+i(y+3)|=\sqrt{x^2+(y+3)^2}$ and $|z-i|=|x+i(y-1)|=\sqrt{x^2+(y-1)^2}$

So the equation becomes $2\sqrt{x^2+(y+3)^2}=\sqrt{x^2+(y-1)^2}$

2. Square both sides

$4\left(x^2+(y+3)^2\right)=x^2+(y-1)^2$

Expand: $4x^2+4(y^2+6y+9)=x^2+(y^2-2y+1)$ $4x^2+4y^2+24y+36=x^2+y^2-2y+1$

Bring all terms to one side: $3x^2+3y^2+26y+35=0$

Divide by $3$: $x^2+y^2+\frac{26}{3}y+\frac{35}{3}=0$

3. Convert to standard circle form

Complete the square in $y$: $y^2+\frac{26}{3}y=\left(y+\frac{13}{3}\right)^2-\frac{169}{9}$

Thus, $x^2+\left(y+\frac{13}{3}\right)^2-\frac{169}{9}+\frac{35}{3}=0$

Now $\frac{35}{3}=\frac{105}{9}$ so $x^2+\left(y+\frac{13}{3}\right)^2-\frac{169}{9}+\frac{105}{9}=0$ $x^2+\left(y+\frac{13}{3}\right)^2=\frac{64}{9}$

Hence this represents a circle with radius $r=\frac{8}{3}$

4. Check options

• A: a circle with radius $\frac{8}{3}$ — Correct
• B: a circle with diameter $\frac{10}{3}$ — false, diameter is $2r=\frac{16}{3}$
• C: an ellipse with major axis $\frac{16}{3}$ — false
• D: an ellipse with minor axis $\frac{16}{9}$ — false

Therefore, the correct option is A.