# Familys of curves in z-plane depending on 1 parameter

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.

-

It's true that

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

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

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 $\pi$.

-
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