Question

If $\left|\frac{\left(z+i\right)}{\left(z-i\right)}\right|=\sqrt{3}$, then the radius of the circle is

• A. $\frac{2}{\sqrt{21}}$
• B. $\frac{1}{\sqrt{21}}$
• C. $\sqrt{3}$
• D. $\sqrt{21}$

Answer 1

The correct option is C

$\sqrt{3}$

Find the radius of the circle.

Let $z=x+iy$.

Now, the equation $\left|\frac{\left(z+i\right)}{\left(z-i\right)}\right|=\sqrt{3}$ can be written as:

$$\begin{gathered} \left|\frac{\left(z+i\right)}{\left(z-i\right)}\right|=\sqrt{3} \\ \Rightarrow \frac{\left|z+i\right|}{\left|z-i\right|}=\sqrt{3} \\ \Rightarrow \left|x+iy+i\right|=\sqrt{3}\left|x+iy-i\right|\left[z=x+iy\right] \\ \Rightarrow \left|x+\left(y+1\right)i\right|=\sqrt{3}\left|x+\left(y-1\right)i\right| \\ \Rightarrow \sqrt{x^2+{\left(y+1\right)}^2}=\sqrt{3}\sqrt{x^2+{\left(y-1\right)}^2}\left[z=x+iy,\left|z\right|=\sqrt{x^2+y^2}\right] \end{gathered}$$

Now square both sides and simplify.

$$\begin{gathered} {\left(\sqrt{x^2+{\left(y+1\right)}^2}\right)}^2={\left(\sqrt{3}\sqrt{x^2+{\left(y-1\right)}^2}\right)}^2 \\ \Rightarrow x^2+y^2+1+2y=3\left(x^2+y^2+1-2y\right) \\ \Rightarrow x^2+y^2+1+2y-3x^2-3y^2-3+6y=0 \\ \Rightarrow -2x^2-2y^2-2+8y=0 \\ \Rightarrow -2\left(x^2+y^2-4y+1\right)=0 \\ \Rightarrow x^2+y^2-4y+1=0 \end{gathered}$$

Now write the equation of circle in standrad form.

$$\begin{gathered} x^2+y^2-4y+1=0 \\ \Rightarrow {\left(x\right)}^2+\left(y^2-4y+4\right)+1-4=0 \\ \Rightarrow {\left(x\right)}^2+{\left(y-2\right)}^2=3 \\ \Rightarrow {\left(x\right)}^2+{\left(y-2\right)}^2={\left(\sqrt{3}\right)}^2 \end{gathered}$$

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

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

Hence, the correct option is C.