I have a problem where I need to find a locus of all points in the complex plane that satisfy $|z-ia|=\lambda|z+ia|$, where $z=x+iy$, and $\lambda>0$.
I know I need to get a circle with the radius $ \left| \dfrac{2\lambda a}{1-\lambda^2} \right|$, centred ant $ z = ia \left[ \dfrac{1+\lambda^2}{1-\lambda^2} \right] $
I tried to express $|z\pm ia|$ in terms of real numbers using $|z \pm ia|=\sqrt{x^2 + (y \pm a)^2}$ and trying to express it in the form of a circle, that is $ (x-\alpha)^2+(y-\beta)^2 = r^2 $, but I just couldn't manage, as the algebra becomes really involved. Maybe someone could help me with the next step in this problem or suggest a different way of approaching it? Not looking for a full answer.
Best Answer
What you are looking for is simply the Apollonius Circle. https://en.wikipedia.org/wiki/Circles_of_Apollonius
If you want to find this explicitly, we have $$x^2+(y-a)^2=\lambda^2x^2+\lambda^2(y+a)^2$$ $$\therefore x^2(1-\lambda^2)+y^2(1-\lambda^2)-2\lambda^2ay-2ay=\lambda^2a^2-a^2 \implies x^2+y^2-\frac{2ay(1+\lambda^2)}{1-\lambda^2}=-a^2$$ Completing the square in the LHS, $$x^2+(y-\frac{a(1+\lambda^2)}{1-\lambda^2})^2=-a^2+(\frac{a(1+\lambda^2)}{1-\lambda^2})^2$$ After a little simplification, we get the desired result $$x^2+(y-\frac{a(1+\lambda^2)}{1-\lambda^2})^2=(\frac{2a\lambda}{1-\lambda^2})^2$$ Hope this helps.