A) \[{{x}^{2}}+{{y}^{2}}-2x-2y-3=0\]
B) \[{{x}^{2}}+{{y}^{2}}+2x-2y-3=0\]
C) \[{{x}^{2}}+{{y}^{2}}+2x+2y-3=0\]
D) \[{{x}^{2}}+{{y}^{2}}-2x+2y-3=0\]
Correct Answer: A
Solution :
Centriod of equilateral triangle is also circumentre of the circle. \[\therefore \] Centre of circle is (1, 1) Radius \[=\sqrt{{{(1+1)}^{2}}+{{(1-2)}^{2}}}=\sqrt{5}\] \[\therefore \] Equation of circle is \[{{(x-1)}^{2}}+{{(y-1)}^{2}}=5\] \[\Rightarrow \,\,{{x}^{2}}+{{y}^{2}}-2x-2y-3=0\]You need to login to perform this action.
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