Question

If $\frac{\left|z-2\right|}{\left|z-3\right|}=2$ represents a circle, then its radius is equal to

• A. $1$
• B. $\frac{1}{3}$
• C. $\frac{3}{4}$
• D. $\frac{2}{3}$

Answer 1

The correct option is D

$\frac{2}{3}$

Explanation for the correct option.

Step 1. Assume the complex number $z$ and simplify the equation.

Let $z=x+iy$ be a complex number. Now the equation $\frac{\left|z-2\right|}{\left|z-3\right|}=2$ becomes:

$$\begin{gathered} \frac{\left|z-2\right|}{\left|z-3\right|}=2 \\ \Rightarrow \left|x+iy-2\right|=2\left|x+iy-3\right| \\ \Rightarrow \left|\left(x-2\right)+iy\right|=2\left|\left(x-3\right)+iy\right| \\ \Rightarrow \sqrt{{\left(x-2\right)}^2+y^2}=2\sqrt{{\left(x-3\right)}^2+y^2} \end{gathered}$$

Now, square both sides.

$$\begin{gathered} {\left(\sqrt{{\left(x-2\right)}^2+y^2}\right)}^2={\left(2\sqrt{{\left(x-3\right)}^2+y^2}\right)}^2 \\ \Rightarrow x^2-4x+4+y^2=4\left(x^2-6x+9+y^2\right) \\ \Rightarrow x^2-4x+4+y^2-4x^2+24x-36-4y^2=0 \\ \Rightarrow -3x^2+20x-3y^2-32=0 \\ \Rightarrow x^2-\frac{20}{3}x+y^2+\frac{32}{3}=0 \end{gathered}$$

Step 2. Write the equation in standard form and find its radius.

The equation $x^2-\frac{20}{3}x+y^2+\frac{32}{3}=0$ can be written in standard form as:

$$\begin{gathered} x^2-\frac{20}{3}x+y^2+\frac{32}{3}=0 \\ \Rightarrow \left(x^2-2·x·\frac{10}{3}+\frac{100}{9}\right)+y^2+\frac{32}{3}-\frac{100}{9}=0 \\ \Rightarrow {\left(x-\frac{10}{3}\right)}^2+y^2=\frac{-96+100}{9} \\ \Rightarrow {\left(x-\frac{10}{3}\right)}^2+y^2=\frac{4}{9} \\ \Rightarrow {\left(x-\frac{10}{3}\right)}^2+y^2={\left(\frac{2}{3}\right)}^2 \end{gathered}$$

On comparing the equation ${\left(x-\frac{10}{3}\right)}^2+y^2={\left(\frac{2}{3}\right)}^2$ with the standard equation of circle ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$, it can be seen that the center of the circle is $\left(\frac{10}{3},0\right)$ and its radius is $\frac{2}{3}$.

So the radius of the circle is $\frac{2}{3}$.

Hence, the correct option is D.