Complex Numbers - Circles

**coobe** · March 22nd 2010, 04:23 AM

Determine the set of solutions for $M = z \epsilon C | |z-3| = 2|z+3|$

Hint: Try to find an equation in form of a general Circle.

Seriously i have no idea.. im quite good in complex numbers, but this is somehow hard for me

**Soroban** · March 22nd 2010, 06:56 AM

Hello, coobe!

Determine the set of solutions for: . $M \;=\;\bigg\{z \in C\;\bigg|\;|z-3| \,=\, 2|z+3|\,\bigg\}$

Hint: Try to find an equation in form of a general circle.

Let $P(x,y)$ be a solution to the equation.

$|z-3|$ is the distance of point $P$ from point $A(3,0).$

$|z+3|$ is the distance of point $P$ from point $B(-3,0).$

We are told that: . $d(PA) \;=\;2\cdot d(PB)$

. . Hence: . $\sqrt{(x-3)^2 + y^2} \;=\;2\sqrt{(x+3)^2 + y^2}$

Square and expand: . $x^2-6x+9 + y^2 \;=\;4(x^2+6x+9 + y^2)$

Simplify: . $3x^2 + 30x + 3y^2 -27 \:=\:0 \quad\Rightarrow\quad x^2 + 10x + y^2 \:=\:-9$

Complete the square: . $x^2 + 10x \:{\color{red}+\: 25} + y^2 \;=\;-9 \:{\color{red}+\: 25 }$

Therefore: . $(x+5)^2 + y^2 \:=\:16$

. . The solution is a circle with center $(-5,0)$ and radius 4.

Note: It is known as the Circle of Apollonius.
.

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