A) \[1\]
B) \[\frac{1}{3}\]
C) \[\frac{3}{4}\]
D) \[\frac{2}{3}\]
Correct Answer: D
Solution :
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\[=\sqrt{{{g}^{2}}+{{f}^{2}}-c}\] \[=\sqrt{\frac{100}{9}-\frac{32}{3}}=\sqrt{\frac{4}{9}}=\frac{2}{3}\]
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