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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.

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1 Answer

up vote 1 down vote accepted

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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Thankyou @joriki. Just may I clarify 1 point: would the $z_{1} $ and $z_{2}$ be on the circle and z in the center, because that's what would be needed for inscribed angle theorem. –  Mykolas Sep 27 '12 at 11:17
sry, I got it. The three of them should be on the circle. Right? –  Mykolas Sep 27 '12 at 11:19
@Mykolas: Yes. The inscribed angle theorem is just needed to show that the angle is always the same; otherwise the centre plays no role. –  joriki Sep 27 '12 at 11:25
Thankyou a lot again @joriki –  Mykolas Sep 27 '12 at 11:26
@Mykolas: You're welcome! –  joriki Sep 27 '12 at 11:32
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