A) \[x+y=0\]
B) \[x-y=\frac{\sqrt{2}}{3}\]
C) \[x-y=0\]
D) None of these
Correct Answer: C
Solution :
We know that the equation of normal to the circle \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] at the point \[({{x}_{1}},\ {{y}_{1}})\] is \[\frac{x}{{{x}_{1}}}-\frac{y}{{{y}_{1}}}=0\]. Therefore\[\frac{x}{1/\sqrt{2}}-\frac{y}{1/\sqrt{2}}=0\Rightarrow x-y=0\].
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