Find the locus of $|z-2i|=3|z+3|$

Question

I got as far as

$$|z-2i|=3|z+3| \Leftrightarrow \\ (\ldots) \Leftrightarrow \\ x^2-y^2+4y-4=9x^2+54x+81-9y^2 \Leftrightarrow \\ x^2-9x^2-y^2+9y^2+4y-4-54x-81=0\Leftrightarrow \\ -8x^2+8y^2+4y-85-54x=0 \Leftrightarrow \\ 8y^2+4y-8x^2-54x-85=0\Leftrightarrow \\ y^2+\frac{1}{2}y-x^2-\frac{27}{4}x=\frac{85}{8}$$

Then I tried to complete the square:

$$y^2+\frac{1}{2}y = 0\Leftrightarrow (y+\frac{1}{4})^2-\frac{1}{16}$$

$$-x^2-\frac{27}{4}x=0 \Leftrightarrow -(x+\frac{27}{8})^2+\frac{729}{64} = 0 \Leftrightarrow (x+\frac{27}{8})^2-\frac{729}{64} = 0$$

And so the equation becomes:

$$(y+\frac{1}{4})^2-\frac{1}{16}+(x+\frac{27}{8})^2-\frac{729}{64}=\frac{85}{8} \Leftrightarrow \\ (y+\frac{1}{4})^2+(x+\frac{27}{8})^2=\frac{85}{8}+\frac{1}{16}+\frac{729}{64} $$

According to my book the center is $-\frac{27}{8}-\frac{1}{4}i$ which I got right but the radius is $\frac{\sqrt{117}}{8}$.

What went wrong?

Answer 1

i have $$|x+iy-2i|=3|x+3+iy|$$ from here we get $$\sqrt{x^2+(y-2)^2}=3\sqrt{(x+3)^2+y^2}$$ can you finish now?

Answer 2

Problem is way up at the top. You wrote $x^2 - y^2 + 4y-4$, when it should have been $x^2 + y^2 - 4y+4$. Similarly, $9x^2 - 9y^2 + 54 x + 81$ should have been $9x^2 + 9y^2 + 54 x + 81$.