Question

If $ax + by = 6$, $bx - ay = 2$ and $x^2 + y^2 = 4$, then the value of $(a^2 + b^2)$ would be:

• A. 10
• B. 2
• C. 4
• D. 5

Hint:

For pairs like $ax+by$ and $bx-ay$, squaring and adding cancels the $xy$ terms, yielding $(a^2+b^2)(x^2+y^2)$ cleanly.

Answer 1

The Correct Option is A

Step 1: Square and add the given linear equations.
\[ (ax+by)^2 + (bx - ay)^2 = \big(a^2x^2 + 2abxy + b^2y^2\big) + \big(b^2x^2 - 2abxy + a^2y^2\big) = (a^2 + b^2)(x^2 + y^2). \] Step 2: Substitute numerical values.
Left side: \(6^2 + 2^2 = 36 + 4 = 40\).
Right side: \((a^2 + b^2)\cdot (x^2+y^2) = (a^2 + b^2)\cdot 4\). \[ 40 = 4(a^2 + b^2) \Rightarrow a^2 + b^2 = 10. \] \[ \boxed{10} \]