Find the radius of the circle in complex plane given by $|z - i| = 3|z +2i|$ where $z$ is a point on the circle. I tried to arrange the following into the standard equation of the circle i.e $|z - z'| = r$ where $r$ is the radius, $z'$ is the centre and $z$ is a point on the circle.
I rearranged the following equation to $|z - i| / |z + 2i| = 3$.
Now I substituted $z = x + iy$ and rationalised the left hand side. After rationalising I could not arrange it to look like the standard form. Please share your solution.
Thank you.
 A: $|z - i| = 3|z +2i|$
$x^2+(y-1)^2=9(x^2+(y+2)^2)$
$x^2+y^2-2y+1=9x^2+9y^2+36y+36$
$8x^2+8y^2+38y+35=0$
$x^2+y^2+\frac{19}{4}y+\frac {35}{8}=0$
$x^2+(y+\frac{19}{8})^2=\frac{81}{64}$
Circle is centered at $(0;-\frac{19}{8})$ with radius $r=\frac{9}{8}$
A: Note that, if $z=x+yi$ with $x,y\in\Bbb R$,\begin{align}|z-i|=3|z+2i|&\iff x^2+(y-1)^2=9x^2+9(y+2)^2\\&\iff8x^2+8y^2+38y=-35\\&\iff8x^2+8\left(y+\frac{19}8\right)^2=\frac{81}8\\&\iff x^2+\left(y+\frac{19}8\right)^2=\left(\frac98\right)^2.\end{align}
A: You can do it as follows:
$$|z-i|^2=9|z+2i|^2\,,$$
and expand this out using complex conjugation and factorize to obtain
$$8|z+\frac{19}{8}i|^2=-35+\frac{19^2}{8}\,,$$
and the radius becomes apparent.
A: If $|z-(a+ib)| = r|z-(c+id)|$ then
$(x-a)^2+(y-b)^2
=r^2((x-c)^2+(y-d)^2)
$
so
$0
=(x^2+y^2)(r^2-1)+2x(a-r^2c)+2y(b-r^2d)+r^2(c^2+d^2)-a^2-b^2\\
=(x^2+y^2)(r^2-1)+2ux+2vy+w
$.
If $r^2=1$ then
$0=2ux+2vy+w
$, a straight line.
If $r^2\ne 1$ then,
if
$(u, v, w)_1
=(u, v, w)/(r^2-1),\\
0
= x^2+2u_1x+y^2+2v_1y+w_1\\
=(x-u_1)^2+(y-v_1)^2+w_1-u_1^2-v_1^2
$
