Find the radius of the circle in complex plane given by $|z - i| = 3|z +2i|$ where $z$ is a point on the circle.

Question

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.

Answer 1

$|z - i| = 3|z +2i|$

$x^2+(y-1)^2=9(x^2+(y+2)^2)$

$x^2+y^2-2y+1=9x^2+9y^2+36y+36$

$8x^2+8y^2+38y+35=0$

$x^2+y^2+\frac{19}{4}y+\frac {35}{8}=0$

$x^2+(y+\frac{19}{8})^2=\frac{81}{64}$

Circle is centered at $(0;-\frac{19}{8})$ with radius $r=\frac{9}{8}$

Answer 2

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}

Answer 3

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.

Answer 4

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 $