Describe the family of curves depending on $C>0$ $\left|\frac{z-z_{1}}{z-z_{2}}\right| = C $ and $arg\frac{z-z_{1}}{z-z_{2}} = C $
What I got:
let $z=x+iy, z_{1}=a+ib, z_{2}=c+id$ $\left|\frac{z-z_{1}}{z-z_{2}}\right| = \frac{(x-a)^{2}+(y-b)^{2}}{(x-c)^{2}+(y-d)^{2}}=C^{2}$ from here: $(1-C^{2})x^{2}-(2a-C^{2}2c)x+(1-C^{2})y^{2}-(2b-C^{2}2d)y=C^{2}d^{2}+C^{2}a^{2}-a^{2}-b^{2}$ which I think is an equation of a circle. Is this correct? For the second question: I am kind of confused... I know that $arg\frac{z-z_{1}}{z-z_{2}}$ represents an angle $z_{1}zz_{2}$, so keeping this constant and equal to C wouldn't just be a point? But I am getting, proceeding similar way like in the first one, an equation of circle, again. But I can't see way it have to be true.