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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.

1 Answers 1

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It's true that

$ \left|\frac{z-z_{1}}{z-z_{2}}\right| = C $

leads to

$ \frac{(x-a)^{2}+(y-b)^{2}}{(x-c)^{2}+(y-d)^{2}}=C^{2} \;, $

but it's wrong to also equate that to

$ \left|\frac{z-z_{1}}{z-z_{2}}\right| $

because it's the square of that. Yes, this is the equation of a circle; so the locus of points with constant ratio of distances to two different points is a circle.

For the second part, you're also right that this leads to a circle equation; see the inscribed angle theorem. However, note that only part of the circle fulfills the original equation; for the other part the angle is shifted by $\pi$.

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    @Mykolas: You're welcome!2012-09-27