If `x,y in R` and satisfy the equation `xy(x^(2)-y^(2))=x^(2)+y^(2)` where `xne0` then the minimum possible value of `x^(2)+y^(2)` is

Answer

To solve the equation \( xy(x^2 - y^2) = x^2 + y^2 \) where \( x, y \in \mathbb{R} \) and \( x \neq 0 \), we will follow these steps:

### Step 1: Substitute Variables
Let \( x = r \cos \theta \) and \( y = r \sin \theta \). This substitution is useful because it allows us to express \( x^2 + y^2 \) in terms of \( r \) and \( \theta \).

### Step 2: Rewrite the Equation
Substituting into the equation gives:
\[ xy(x^2 - y^2) = (r \cos \theta)(r \sin \theta)((r \cos \theta)^2 - (r \sin \theta)^2) \]
This simplifies to:
\[ r^2 \cos \theta \sin \theta (r^2 \cos^2 \theta - r^2 \sin^2 \theta) = r^2 \cos^2 \theta + r^2 \sin^2 \theta \]
Factoring out \( r^2 \) from both sides, we have:
\[ r^2 \cos \theta \sin \theta (r^2 (\cos^2 \theta - \sin^2 \theta)) = r^2 \]

### Step 3: Simplify the Equation
Since \( r^2 \) is not zero (as \( x \neq 0 \)), we can divide both sides by \( r^2 \):
\[ \cos \theta \sin \theta (\cos^2 \theta - \sin^2 \theta) = 1 \]

### Step 4: Use Trigonometric Identities
Recall that \( \sin 2\theta = 2 \sin \theta \cos \theta \) and \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \). Thus, we can rewrite the equation as:
\[ \frac{1}{2} \sin 2\theta \cos 2\theta = 1 \]
Multiplying both sides by 2 gives:
\[ \sin 2\theta \cos 2\theta = 2 \]

### Step 5: Further Simplification
Using the identity \( \sin 4\theta = 2 \sin 2\theta \cos 2\theta \), we can rewrite the equation as:
\[ \frac{1}{2} \sin 4\theta = 2 \]
Thus, we have:
\[ \sin 4\theta = 4 \]
However, since the sine function has a maximum value of 1, we need to adjust our approach.

### Step 6: Find Minimum Value of \( r^2 \)
From our previous steps, we derived:
\[ r^2 = \frac{4}{\sin 4\theta} \]
To minimize \( r^2 \), we need to maximize \( \sin 4\theta \). The maximum value of \( \sin 4\theta \) is 1, occurring when \( 4\theta = \frac{\pi}{2} + 2k\pi \) for \( k \in \mathbb{Z} \).

### Step 7: Calculate Minimum Value
Thus, the minimum value of \( r^2 \) occurs when \( \sin 4\theta = 1 \):
\[ r^2 = \frac{4}{1} = 4 \]
Therefore, the minimum possible value of \( x^2 + y^2 \) is:
\[ \boxed{4} \]