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.

Question

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?

Answer 1

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.

Answer 2

$|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.

Answer 3

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

$$z_c=z_1+\frac{1}{1-c^2}(z_2-z_1)$$

Answer 4

Let $z=x+iy$, $z_1=x_1+iy_1$, $z_2=x_2+iy_2$. Then the equation becomes: \begin{eqnarray*} |z-z_1| &=& c|z-z_2|\\ |z-z_1|^2 &=& c^2|z-z_2|^2\\ (x-x_1)^2+(y-y_1)^2 &=& c^2(x-x_2)^2+c^2(y-y_2)^2\\ x^2-2xx_1+x_1^2+y^2-2yy_1+y_1^2 &=& c^2x^2-2c^2xx_2+c^2x_2^2+c^2y^2-2c^2yy_2+c^2y_2^2\\ x^2(1-c^2)-2x(x_1-c^2x_2)+y^2(1-c^2)-2y(y_1-c^2y_2) &=& c^2(x_2^2+y_2^2)-(x_1^2+y_1^2)\\ x^2-2x\left(\frac{x_1-c^2x_2}{1-c^2}\right)+y^2-2y\left(\frac{y_1-c^2y_2}{1-c^2}\right) &=& \left(\frac{c^2(x_2^2+y_2^2)}{1-c^2}\right)-\left(\frac{x_1^2+y_1^2}{1-c^2}\right)\\ x^2-2x\left(\frac{x_1-c^2x_2}{1-c^2}\right)+\left(\frac{x_1-c^2x_2}{1-c^2}\right)^2+y^2-2y\left(\frac{y_1-c^2y_2}{1-c^2}\right)+\left(\frac{y_1-c^2y_2}{1-c^2}\right)^2 &=& \left(\frac{c^2(x_2^2+y_2^2)}{1-c^2}\right)-\left(\frac{x_1^2+y_1^2}{1-c^2}\right)+\left(\frac{x_1-c^2x_2}{1-c^2}\right)^2+\left(\frac{y_1-c^2y_2}{1-c^2}\right)^2\\ \left(x-\left(\frac{x_1-c^2x_2}{1-c^2}\right)\right)^2+\left(y-\left(\frac{y_1-c^2y_2}{1-c^2}\right)\right)^2 &=& \left(\frac{c^2(x_2^2+y_2^2)}{1-c^2}\right)-\left(\frac{x_1^2+y_1^2}{1-c^2}\right)+\left(\frac{x_1-c^2x_2}{1-c^2}\right)^2+\left(\frac{y_1-c^2y_2}{1-c^2}\right)^2\\ \left(x-\left(\frac{x_1-c^2x_2}{1-c^2}\right)\right)^2+\left(y-\left(\frac{y_1-c^2y_2}{1-c^2}\right)\right)^2 &=& \frac{c^2(x_2-x_1)^2+c^2(y_2-y_1)^2}{(1-c^2)^2} \end{eqnarray*} So the center appears to be $z_c=\frac{z_1-c^2z_2}{1-c^2}=z_1+\frac{c^2(z_1-z_2)}{1-c^2}$ with radius $$ r=\sqrt{\frac{c^2(x_2-x_1)^2+c^2(y_2-y_1)^2}{(1-c^2)^2}}=\frac{c}{|1-c^2|}|z_2-z_1| $$ Recall that $c>0$, $c\neq 1$.

Funny; there are several answers to this question and each are different!

Answer 5

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 $\frac{ | z_1-z_2 |c}{|1-c^2|}$.