If $c >0$, $c≠ 1$, $z_1≠z_2$, prove that $\frac{|z-z_1 |}{|z-z_2 | }=c$ represents a circle. Find its center and radius. I am studying complex analysis, and I was solving some questions regarding the subject when I stumbled upon this question. The equation above represents a circle, where $c$ is an arbitrary complex number. I don't get how it represents a circle, I tried proving it but I failed. Any help in finding the radius and center and proving the equation?
 A: $|z-z_{0}| = c $ then the center of the circle is $z_{0}$ and radius c.
$| \frac{z-z_{1}}{z-z_{2}}|=c$, assumed $c>0$
$\Rightarrow |z-z_{1}|=|(z-z_{2})|c$, assumed $c>0$
$\Rightarrow |z-z_{1}|-|(z-z_{2})|c=0$, assumed $c>0$
$\Rightarrow |z-cz-(z_{1}-z_{2}c)|=0 $ consider $|a|-|b|=0 \Rightarrow |a-b|=0$ where $a,b>0$ as well as assumed few additional condition.
$\Rightarrow |z(1-c)-(z_{1}-z_{2}c)|=0 $
$\Rightarrow |z(1-c)|=|(z_{1}-z_{2}c)| $consider $|a-b|=0 \Rightarrow |a|-|b|=0$ where $a,b>0$ as well as assumed few additional condition.
$\Rightarrow |z-0|=\frac{|(z_{1}-z_{2}c)|}{|1-c|}=c_{1}(say) $
Then the center of the circle is (0,0) and radius is $\frac{|(z_{1}-z_{2}c)|}{|1-c|}=c_{1}$ where $0<c<1$ is also considered.
A: Here's a geometric way to answer this question. Everything that follows refers to the following figure where we identify $A=z_1$, $B=z_2$ and we have used $$\frac{|z-z_2|}{|z-z_1|}=c$$
It is clear from the definition of the circle by Apollonius that the given equation is a circle, since it restricts the points $z$ of the plane to those whose distances from two given foci, situated at $z_1,z_2$. 
Now that we know that the locus in question is a circle, consider the points $C,D$ of the figure. For those points it is true that:
$$\frac{|AC|}{|BC|}=\frac{|AD|}{|BD|}=c$$
However it is true $|AC|+|BC|=|z_2-z_1|$. We conclude that
$$|AC|=x =\frac{|z_2-z_1|}{1+c}~,~|BC|=y=\frac{c|z_2-z_1|}{1+c}$$
Furthermore it is true that $|AD|=2R+|AC|$, while $|BD|=2R-|BC|$ and using the ratio of these two distances we can see that the radius of the circle is (consider both cases c>1, and c<1):
$$\frac{2R+x}{2R-y}=c\iff R=\frac{c}{|1-c^2|}|z_2-z_1|$$
Now it is obvious that the center of the circle is in the same line as the two foci, and thus there has to be a constant $\lambda\in \mathbb{R}$ such that
$$z_c=z_1+\lambda(z_2-z_1)$$
Note that in the figure $\lambda>0$ and therefore $c<1$. All we need to know is the distance $|z_c-z_1|$. We can find it in terms of known quantities because $|z_c-z_1|=x+R=\frac{|z_1-z_2|}{|1-c^2|}$ (again we need to separate the cases $c>1$ and $c<1$). Now since
$$|\lambda|=\frac{|z_c-z_1|}{|z_2-z_1|}\Rightarrow |\lambda|=\frac{1}{|1-c^2|}$$
we conclude that the position of the center is 
$$z_c=z_1+\frac{1}{1-c^2}(z_2-z_1)$$
A: Assuming $c$ is a real number, let  $$z=x+yi,z_1=x_1+y_1i,z_2=x_2+y_2i$$ where all the $x$'s and $y$'s are real.  Multiply both sides of your given equation by $z-z_2$ which gives $$(x-x_1)+(y-y_1)i=c((x-x_2)+(y-y_2)i)$$ Take the conjugate on both sides and multiply.Thus $$(x-x_1)^2+(y-y_1)^2=c^2((x-x_2)^2+(y-y_2)^2)$$ Elementary algebra will put this equation into the standard form of a circle.
A: Rewrite the given equation $\frac{|z-z_1 |}{|z-z_2 | }=c$ as
$$(z-z_1)(\bar z-\bar z_1 )=c^2 (z-z_2 )(\bar z-\bar z_2 ) $$
which leads to
$$|z |^2-\frac{ z_1-z_2 c^2}{1-c^2}\bar z -\frac{ \bar z_1-\bar z_2c^2}{1-c^2}z
= \frac{ |z_2|^2 c^2 -|z_1|^2}{1-c^2}$$
or explicitly in the form of the circle equation
$$\bigg|z -\frac{ z_1-z_2 c^2}{1-c^2}\bigg|^2
= \left(\frac{ | z_1-z_2 | c}{1-c^2}\right)^2$$
Thus, the center is $ \frac{ z_1-z_2 c^2}{1-c^2}$ and the radius 
$\frac{ | z_1-z_2 |c}{|1-c^2|}$.
