Question

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

A

$\frac{2}{\sqrt{21}}$

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B

$\frac{1}{\sqrt{21}}$

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C

$\sqrt{3}$

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D

$\sqrt{21}$

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Solution

## 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:$\left|\frac{\left(z+i\right)}{\left(z-i\right)}\right|=\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒\frac{\left|z+i\right|}{\left|z-i\right|}=\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒\left|x+iy+i\right|=\sqrt{3}\left|x+iy-i\right|\left[z=x+iy\right]\phantom{\rule{0ex}{0ex}}⇒\left|x+\left(y+1\right)i\right|=\sqrt{3}\left|x+\left(y-1\right)i\right|\phantom{\rule{0ex}{0ex}}⇒\sqrt{{x}^{2}+{\left(y+1\right)}^{2}}=\sqrt{3}\sqrt{{x}^{2}+{\left(y-1\right)}^{2}}\left[z=x+iy,\left|z\right|=\sqrt{{x}^{2}+{y}^{2}}\right]$Now square both sides and simplify.${\left(\sqrt{{x}^{2}+{\left(y+1\right)}^{2}}\right)}^{2}={\left(\sqrt{3}\sqrt{{x}^{2}+{\left(y-1\right)}^{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}+1+2y=3\left({x}^{2}+{y}^{2}+1-2y\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}+1+2y-3{x}^{2}-3{y}^{2}-3+6y=0\phantom{\rule{0ex}{0ex}}⇒-2{x}^{2}-2{y}^{2}-2+8y=0\phantom{\rule{0ex}{0ex}}⇒-2\left({x}^{2}+{y}^{2}-4y+1\right)=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-4y+1=0$Now write the equation of circle in standrad form.${x}^{2}+{y}^{2}-4y+1=0\phantom{\rule{0ex}{0ex}}⇒{\left(x\right)}^{2}+\left({y}^{2}-4y+4\right)+1-4=0\phantom{\rule{0ex}{0ex}}⇒{\left(x\right)}^{2}+{\left(y-2\right)}^{2}=3\phantom{\rule{0ex}{0ex}}⇒{\left(x\right)}^{2}+{\left(y-2\right)}^{2}={\left(\sqrt{3}\right)}^{2}$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.

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