A) \[{{x}^{2}}+{{y}^{2}}+2x-2y=0\]
B) \[{{x}^{2}}+{{y}^{2}}+2x-2y+1=0\]
C) \[{{x}^{2}}+{{y}^{2}}+2x-2y-1=0\]
D) \[{{x}^{2}}+{{y}^{2}}-2x+2y+1=0\]
Correct Answer: B
Solution :
By plotting, we get the centre of the circle, which is inscribed in the square OABC is \[(-1,1)\] and radius is 1. \[\therefore \] Equation of circle is \[{{(x+1)}^{2}}+{{(y-1)}^{2}}={{1}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}+2x-2y+1=0\]You need to login to perform this action.
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