Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $\pi$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.