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: \((a^2 + b^2)\cdot (x^2+y^2) = (a^2 + b^2)\cdot 4\). \[ 40 = 4(a^2 + b^2) \Rightarrow a^2 + b^2 = 10. \] \[ \boxed{10} \]

Answer 2

A neat alternative uses complex numbers: treat \(a+ib\) and \(x+iy\) as complex numbers and combine the two given linear equations into one complex equation.

Forming the complex product. Consider \( (a+ib)(x-iy) = (ax+by) + i(bx-ay) \), since expanding gives \( ax -iay+ibx -i^2by = (ax+by)+i(bx-ay) \). Substituting the given values \(ax+by=6\) and \(bx-ay=2\): \[ (a+ib)(x-iy) = 6+2i. \]

Taking magnitudes. Taking the modulus of both sides and using \(|z_1z_2|=|z_1||z_2|\): \[ |a+ib|\cdot|x-iy| = |6+2i| = \sqrt{36+4}=\sqrt{40}. \] Since \(|x-iy|=\sqrt{x^2+y^2}=\sqrt4=2\), \[ |a+ib| = \frac{\sqrt{40}}{2}=\sqrt{10} \ \Rightarrow\ a^2+b^2 = 10. \]

1. Option 10: Matches \( |a+ib|^2=10 \) exactly as derived from the modulus relation.
2. Option 2: Would require \(|6+2i|^2\) to equal \(2\times4=8\), but \(6^2+2^2=40\neq8\); ruled out.
3. Option 4: Would give \(4\times4=16\neq40\); ruled out.
4. Option 5: Would give \(5\times4=20\neq40\); ruled out.

Only \(a^2+b^2=10\) is consistent with \(|a+ib|^2\cdot|x+iy|^2=|6+2i|^2=40\) together with \(x^2+y^2=4\).

So the correct answer is \(10\).