A) straight line
B) circle
C) parabola
D) None of these
Correct Answer: B
Solution :
Given, \[z=x+iy\] and \[\arg \left( \frac{z-2}{z+2} \right)=\frac{\pi }{6}\] \[\Rightarrow \] \[{{\tan }^{-1}}\left( \frac{z-2}{z+2} \right)=\frac{\pi }{6}\] \[\Rightarrow \] \[\frac{z-2}{z+2}=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}\] \[\Rightarrow \] \[\frac{(x+iy)-2}{(x+iy)+2}=\frac{1}{\sqrt{3}}\] \[\Rightarrow \]\[\frac{(x-2)+iy}{(x+2)+iy}\times \frac{(x+2)-iy}{(x+2)-iy}=\frac{1}{\sqrt{3}}\] \[\Rightarrow \]\[\frac{({{x}^{2}}-4)-i(x-2)y+i(x+2)y-{{i}^{2}}{{y}^{2}}}{{{(x+2)}^{2}}-{{(iy)}^{2}}}\] \[=\frac{1}{\sqrt{3}}\] \[\Rightarrow \]\[\frac{{{x}^{2}}-4+{{y}^{2}}+i(-xy+2y+xy+2y)}{{{(x+2)}^{2}}+{{y}^{2}}}=\frac{1}{\sqrt{3}}\] On comparing the real parts, \[\frac{{{x}^{2}}-4+{{y}^{2}}}{{{(x+2)}^{2}}+{{y}^{2}}}=\frac{1}{\sqrt{3}}\] \[\Rightarrow \] \[\sqrt{3}{{x}^{2}}+\sqrt{3}{{y}^{2}}-4\sqrt{3}={{x}^{2}}+4x+4+{{y}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}(\sqrt{3}-1)+{{y}^{2}}(\sqrt{3}-1)-4x=4+4\sqrt{3}\] which is the equation of circle.
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