If `(x^(2)+y^(2)) (p^(2) +q^(2)) =(xp +yq)^(2) ` then

Options

A. `xy=pq`
B. `px=yq`
C. `xq= yp`
D. None of these

Answer

To solve the equation \( (x^2 + y^2)(p^2 + q^2) = (xp + yq)^2 \), we will follow these steps:

### Step 1: Expand the left-hand side
We start with the left-hand side of the equation:
\[ (x^2 + y^2)(p^2 + q^2) \]
Using the distributive property (also known as the FOIL method), we can expand this:
\[ = x^2p^2 + x^2q^2 + y^2p^2 + y^2q^2 \]

### Step 2: Expand the right-hand side
Now, we expand the right-hand side:
\[ (xp + yq)^2 \]
Using the square of a binomial formula \( (a + b)^2 = a^2 + 2ab + b^2 \), we get:
\[ = (xp)^2 + 2(xp)(yq) + (yq)^2 \]
This simplifies to:
\[ = x^2p^2 + 2xy(pq) + y^2q^2 \]

### Step 3: Set the two expanded forms equal to each other
Now we have:
\[ x^2p^2 + x^2q^2 + y^2p^2 + y^2q^2 = x^2p^2 + 2xy(pq) + y^2q^2 \]

### Step 4: Cancel out the common terms
We can cancel \( x^2p^2 \) and \( y^2q^2 \) from both sides:
\[ x^2q^2 + y^2p^2 = 2xy(pq) \]

### Step 5: Rearrange the equation
Rearranging gives us:
\[ x^2q^2 + y^2p^2 - 2xy(pq) = 0 \]

### Step 6: Recognize the identity
The left-hand side can be recognized as a perfect square:
\[ (xq - yp)^2 = 0 \]
This implies:
\[ xq - yp = 0 \]

### Step 7: Solve for the relationship between x, y, p, and q
From \( xq - yp = 0 \), we can conclude:
\[ xq = yp \]

### Conclusion
Thus, the relationship derived from the original equation is:
\[ xq = yp \]