A) \[\frac{2}{\sqrt{3}}{{\left( \frac{\tau }{Bi} \right)}^{1/2}}\]
B) \[\frac{1}{\sqrt{3}}\frac{\tau }{Bi}\]
C) \[{{m}_{1}}\]
D) \[{{m}_{2}}\]
Correct Answer: D
Solution :
Let the equation of the circle be\[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\]which passes through (0,0) and (1, 0). \[\therefore \] c = 0and 1 + 2g + c = 0 \[\Rightarrow \]\[c=0,g=-\frac{1}{2}\] The circle \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\]touch the circle\[{{x}^{2}}+{{y}^{2}}=9\] \[\therefore \]\[\sqrt{{{g}^{2}}+{{f}^{2}}}=3\pm \sqrt{{{g}^{2}}+{{f}^{2}}-c}\] \[[\because {{C}_{1}}{{C}_{2}}={{r}_{1}}\pm {{r}_{2}}]\] \[\Rightarrow \]\[\sqrt{\frac{1}{4}+{{f}^{2}}}=3\pm \sqrt{\frac{1}{4}+{{f}^{2}}}\]\[\Rightarrow \]\[3=2\sqrt{\frac{1}{4}+{{f}^{2}}}\] \[\Rightarrow \]\[{{f}^{2}}=\frac{9}{4}-\frac{1}{4}=2\]\[\Rightarrow \]\[f=\pm \sqrt{2}\] Hence, the coordinates of the centre are \[\left( \frac{1}{2},\pm \sqrt{2} \right)\]
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