A) 1
B) \[1/3\]
C) \[3/4\]
D) \[2/3\]
Correct Answer: D
Solution :
Given,\[\frac{|z-2|}{|z-3|}=2\] Þ \[\sqrt{{{(x-2)}^{2}}+{{y}^{2}}}=2\sqrt{{{(x-3)}^{2}}+{{y}^{2}}}\] Þ \[{{(x-2)}^{2}}+{{y}^{2}}=4[{{(x-3)}^{2}}+{{y}^{2}}]\] Þ \[{{x}^{2}}+{{y}^{2}}+4-4x=4{{x}^{2}}+4{{y}^{2}}+36-24x\] Þ \[3{{x}^{2}}+3{{y}^{2}}-20x+32=0\] or \[{{x}^{2}}+{{y}^{2}}-\frac{20}{3}x+\frac{32}{3}=0\] .....(i) We know that, standard equation of circle, \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] .....(ii) Comparison of (i) from(ii) Þ \[2g=-\frac{20}{3}\,\Rightarrow g=-\frac{10}{3},f=0,c=\frac{32}{3}\] 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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