<s>[INST] Find the Cartesian equation of the locus described by $|z+2-7i| = 2|z-10+2i|$ 
Find the Cartesian equation of the locus described by $|z+2-7i| =
 2|z-10+2i|$ Write your answer in the form $(x+a)^2+(y+b)^2=k$.

This was a question from my end of year exams just gone and I'm unsure as to where I have gone wrong :(. If anyone could point me in the right direction I'd very much appreciate that.
Here was my working:
Let $z = x+iy$
$|z|$ = $\sqrt{x^2+y^2}$
$$|z+2-7i| = 2|z-10+2i| \\
\implies |(x+iy)+2-7i| = 2|(x+iy)-10+2i| \\  
\implies |(x+2)(y-7y)i| = 2|(x-10)+(y+2)i| \\ 
\implies \sqrt{(x+2)^2+(y-7)^2i^2} = 2\sqrt{(x-10)^2+(y+2)^2i^2}  \\
\implies (x+2)^2-((y-7^2) = 2((x-10)^2-(y+2)^2)$$
$$(x^2+4x+4)-(y^2-14y+49) = 2((x^2-20x+100)-(y^2+2y+4)) \\
\implies x^2+4x+4-y^2+14y-49 = 2x^2-40x+200-2y^2-4y-8 \\
\implies -237 = x^2-44x-y^2-18y\\
\implies x^2-44x-y^2-18 = -237$$
 [/INST] Consider $|z+a+bi|=2|z+c+di|$. Squaring gives
$$(x+a)^2+(y+b)^2=4(x+c)^2+4(y+d)^2$$
that is
$$3(x^2+y^2)+(8c-2a)x+(8d-2b)y+4c^2+4d^2-a^2-b^2=0.$$
In general this is a circle. Now put in your values of $a,\ldots,d$.
</s>