The locus $\left|\dfrac{z-2}{z+3i}\right|=4$ simplifies to which equation?

1. Use $\left|\dfrac{z_1}{z_2}\right|=\dfrac{|z_1|}{|z_2|}$, so $|z-2|=4|z+3i|$. 2. Put $z=x+iy$: $\sqrt{(x-2)^2+y^2}=4\sqrt{x^2+(y+3)^2}$. 3. Square both sides: $(x-2)^2+y^2=16\left[x^2+(y+3)^2\right]$. 4. Expanding gives $x^2-4x+4+y^2=16x^2+16y^2+96y+144$. 5. Collecting terms: $15x^2+15y^2+4x+96y+140=0$. Equal coefficients on $x^2$ and $y^2$ confirm a circle; the version with $-15y^2$ breaks that and would be a hyperbola. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/2, Question 4: Complex Numbers_