If `x ^(2) + y ^(2) + (1)/( x ^(2)) + (1)/ y ^(2) = 4,` then the value of `x ^(2) + y ^(2) ` is :

Options

A. a. 2
B. b. 4
C. c. 8
D. d. 16

Answer

To solve the equation \( x^2 + y^2 + \frac{1}{x^2} + \frac{1}{y^2} = 4 \), we can start by letting \( a = x^2 \) and \( b = y^2 \). Thus, we can rewrite the equation as:

\[ a + b + \frac{1}{a} + \frac{1}{b} = 4 \]

Next, we can combine the terms involving \( a \) and \( b \):

\[ a + b + \frac{1}{a} + \frac{1}{b} = a + b + \frac{a + b}{ab} \]

This can be simplified to:

\[ a + b + \frac{a + b}{ab} = 4 \]

Let \( s = a + b \) and \( p = ab \). Then we can rewrite the equation as:

\[ s + \frac{s}{p} = 4 \]

Multiplying through by \( p \) gives:

\[ sp + s = 4p \]

Rearranging this yields:

\[ sp - 4p + s = 0 \]

Factoring out \( s \):

\[ s(p + 1) = 4p \]

Thus, we have:

\[ s = \frac{4p}{p + 1} \]

Now, we need to find the minimum value of \( s \). To do this, we can analyze the function \( s(p) = \frac{4p}{p + 1} \).

To find the critical points, we can differentiate \( s \) with respect to \( p \):

\[ s' = \frac{(p + 1)(4) - 4p(1)}{(p + 1)^2} = \frac{4}{(p + 1)^2} \]

Setting \( s' = 0 \) does not yield any critical points since the numerator is a constant. However, we can analyze the behavior of \( s \) as \( p \) approaches certain limits.

As \( p \to 0 \):

\[ s \to 0 \]

As \( p \to \infty \):

\[ s \to 4 \]

Thus, the maximum value of \( s \) occurs when \( p \) is large, and we can also check if \( s = 2 \) is achievable.

Now, substituting \( s = 2 \) into the original equation:

\[ 2 + \frac{2}{p} = 4 \implies \frac{2}{p} = 2 \implies p = 1 \]

Thus, we have \( a + b = 2 \) and \( ab = 1 \).

The values of \( a \) and \( b \) can be found by solving the quadratic equation:

\[ t^2 - st + p = 0 \implies t^2 - 2t + 1 = 0 \]

This gives:

\[ (t - 1)^2 = 0 \implies t = 1 \]

Thus, \( a = 1 \) and \( b = 1 \), which means:

\[ x^2 + y^2 = a + b = 1 + 1 = 2 \]

Therefore, the value of \( x^2 + y^2 \) is:

\[ \boxed{2} \]