Question

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

Answer 1

Find the required value.

It is given that, $x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4$

$$\begin{gathered} \Rightarrow x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}-4=0 \\ \Rightarrow x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}-2-2=0 \\ \Rightarrow \left(x^2+\frac{1}{x^2}-2\right)+\left(y^2+\frac{1}{y^2}-2\right)=0 \\ \Rightarrow {\left(x-\frac{1}{x}\right)}^2+{\left(y-\frac{1}{y}\right)}^2=0 \end{gathered}$$

Since, both the terms in the above equation must be greater than or equal to zero.

$$\begin{gathered} \Rightarrow \left(x-\frac{1}{x}\right)=0,\left(y-\frac{1}{y}\right)=0 \\ \Rightarrow x=\frac{1}{x},y=\frac{1}{y} \\ \Rightarrow x^2=1,y^2=1 \end{gathered}$$

Therefore, $x^2+y^2=1+1$.

Hence, the value of $x^2+y^2$ is $2$.