A) \[{{x}^{2}}+{{y}^{2}}-2x+2y=0\]
B) \[{{x}^{2}}+{{y}^{2}}+2x-2y=0\]
C) \[{{x}^{2}}+{{y}^{2}}+2x+2y=0\]
D) \[{{x}^{2}}+{{y}^{2}}-2x-2y=0\]
Correct Answer: C
Solution :
Since the circle passes through (0, 0), hence\[c=0\]. Also \[2\sqrt{{{g}^{2}}-c}=2\Rightarrow g=1\] and \[2\sqrt{{{f}^{2}}-c}=2\Rightarrow f=1\]. Hence radius is \[\sqrt{2}\] and centre is\[(-1,\ -1)\]. Therefore, the required equation is\[{{x}^{2}}+{{y}^{2}}+2x+2y=0\]. Trick: Obviously the centre of circle lies in III quadrant, which is given by (c).
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