If $x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4,$ then the value of $x^2+y^2$ is

The correct option is A 2
Since the question contains variables, we can assume any value for x and y.
Let, x = y = 1. Then,
$x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=1+1+1+1=4=4,$ which is true as per given equation.
$\therefore$ $x^2+y^2=1+1=2.$
Hence, option (a) is correct answer.

Alternatively:
$x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}-4=0\Rightarrow x^2+\frac{1}{x^2}-2+y^2+\frac{1}{y^2}-2=0\Rightarrow {(x-\frac{1}{x})}^2+{(y-\frac{1}{y})}^2=0\Rightarrow x-\frac{1}{x}=0\Rightarrow x^2-1=0\Rightarrow x=1,-1Similary,y=1,-1\therefore x^2+y^2=1+1=2$