Question : If $x^{2}+y^{2}+\frac{1}{x^{2}}+\frac{1}{y^{2}}=4$, then the value of $x^{2}+y^{2}$ is:
Option 1: 2
Option 2: 4
Option 3: 8
Option 4: 16
Correct Answer: 2
Solution : $x^{2}+y^{2}+\frac{1}{x^{2}}+\frac{1}{y^{2}}-4=0$ $⇒x^{2}+\frac{1}{x^{2}}-2+y^{2}+\frac{1}{y^{2}}-2=0$ $⇒(x-\frac{1}{x})^2+(y-\frac{1}{y})^2 = 0$ So, $x-\frac{1}{x}=0$ and $y-\frac{1}{y}=0$ ⇒ $x^2-1=0$ and $y^2-1=0$ ⇒ $x^2 = 1$ and $y^2 = 1$ $\therefore x^{2}+y^{2}= 1 + 1 = 2$ Hence, the correct answer is 2.
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