Finding the locus of a $z=x+iy$ and its radius if $|z-1|=2|z+2-3i|$ 
If the point $P$ in the complex plane corresponds to the complex number $z=x+iy$ show that if $|z-1|=2|z+2-3i|$ then the locus of $P$ is a circle centre at $-3+4i$, and find the radius of the circle.

Putting them into cartesian equations, we have:
$$
(x-1)^2+y^2=2[(x+2)^2+(y-3)^2]\\
x^2-2x+1+y^2=2x^2+8x+8+2y^2-12y+18\\
-x^2-10x-y^2+12y=25
$$
So it seems as if the centre must be $(-5,6)$ or $-5+6i$ with a radius of $5$, but that doesn't match the correct answer ($2\sqrt2)$. Am I doing something wrong?

With thanks to Anurag A:
$$
(x-1)^2+y^2=4[(x+2)^2+(y-3)^2]\\
0=3x^2+18x+3y^2-24y+51\\
-(x+3)^2-(y-4)^2=-8\\
$$
The centre is therefore $(-3,4)$ or $-3+4i$ and the radius is $\sqrt8$ or $2\sqrt2$.
