Question

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

• A. 2
• B. 4
• C. 8
• D. 16

Answer 1

The correct option is A 2

$x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4$

$\Rightarrow x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}-4=0$

$x^2+\frac{1}{x^2}+y^2+\frac{1}{y^2}-2-2=0$

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

${\left(x-\frac{1}{x}\right)}^2+{\left(y-\frac{1}{y}\right)}^2=0$

$\Rightarrow \left(x-\frac{1}{x}\right)=0$ $\&$ $\left(y-\frac{1}{y}\right)=0$ ( Since ${\left(x-\frac{1}{x}\right)}^2$ $\&$ ${\left(y-\frac{1}{y}\right)}^2$

(both must be greater than or equal to zero )

$\Rightarrow x=\frac{1}{x}$ $\&$ $y=\frac{1}{y}$

$\Rightarrow x^2=1$ $\&$ $y^2=1$

$\Rightarrow x^2+y^2=1+1=2.$

Hence, the answer is $2.$

OR

$x^2+y^2\frac{1}{x^2}+\frac{1}{y^2}=4$
Putting x = y = 1, we get
LHS = 1 + 1 $\frac{1}{1}+\frac{1}{1}=4$
RHS = 4
LHS = RHS
$x^2+y^2=(1{)}^2+(1{)}^2=2$