A) \[\frac{9}{8}\]
B) \[\frac{81}{64}\]
C) \[\frac{9}{64}\]
D) 9
Correct Answer: C
Solution :
\[|z{{|}^{2}}\,=9|z-1{{|}^{2}}\] \[{{x}^{2}}+{{y}^{2}}=9[{{(x-1)}^{2}}+{{y}^{2}}]\] \[{{x}^{2}}+{{y}^{2}}\,=9({{x}^{2}}+{{y}^{2}}-2x+1)\] \[8{{x}^{2}}+8{{y}^{2}}-18x+9=0\] \[{{\left( x-\frac{9}{8} \right)}^{2}}+{{y}^{2}}=\frac{9}{64}\] \[\therefore \] Area \[=\frac{9\pi }{64}\] and\[e=0\] \[\therefore \,\,a=\frac{9}{64}\]You need to login to perform this action.
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