A) \[{{x}^{2}}+{{y}^{2}}-4y=0\]
B) \[{{x}^{2}}+{{y}^{2}}+4x=0\]
C) \[{{x}^{2}}+{{y}^{2}}-4y=12\]
D) \[{{x}^{2}}+{{y}^{2}}+4x=12\]
Correct Answer: C
Solution :
Solving \[y=0\] and\[y+\sqrt{3}x=6\], we get\[(2\sqrt{3},\,0)\], only option (c) satisfies the co-ordinate.
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