Question

If $ \frac{|z-2|}{|z-3|}=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

Answer: (D)

Given, $ \frac{|z-2|}{|z-3|}=2 $
$ \Rightarrow $ $ |z-2|=2|z-3| $
$ \Rightarrow $ $ \sqrt{{{(x-2)}^{2}}+{{y}^{2}}}=2\sqrt{{{(x-3)}^{2}}+{{y}^{2}}} $
$ \Rightarrow $ $ {{(x-2)}^{2}}+{{y}^{2}}=4[{{(x-3)}^{2}}+{{y}^{2}}] $ (on squaring both sides)
$ \Rightarrow $ $ {{x}^{2}}+{{y}^{2}}-4x+4=4{{x}^{2}}+4{{y}^{2}}+36-24x $
$ \Rightarrow $ $ 3{{x}^{2}}+3{{y}^{2}}-20x+32=0 $
or $ {{x}^{2}}+{{y}^{2}}-\frac{20}{3}x+\frac{32}{3}=0 $ ... (i)
We know that, standard equation of a circle is
$ {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 $ ... (ii)
On comparing Eqs. (i) and (ii), we get
$ 2g=\frac{-20}{3}\Rightarrow g=\frac{-10}{3},\,\,f=0,\,\,c=\frac{32}{3} $
Hence, radius $ =\sqrt{{{g}^{2}}+{{f}^{2}}-c} $
$ =\sqrt{\frac{100}{9}-\frac{32}{3}}=\sqrt{\frac{4}{9}}=\frac{2}{3} $