Question

If ax+by=3, bx-ay=4 and $x^2+y^2=1$ then the value of $a^2+b^2$ is

• A. 25
• B. 26
• C. 27
• D. 28

Answer 1

The correct option is A 25

Given $x^2+y^2=1$ and

ax+by=3 and bx-ay=4 Then square both equation and added

$(ax+by{)}^2+(bx-ay{)}^2=(3{)}^2+(4{)}^2$

$\Rightarrow a^2{x}^2+b^2{y}^2+2abxy+b^2{x}^2+a^2{y}^2-2abxy=9+16$

$\Rightarrow a^2{x}^2+b^2{x}^2+a^2{y}^2+b^2{y}^2=25$

$\Rightarrow (a^2+b^2)(x^2+y^2)=25$

$\Rightarrow (a^2+b^2)\times 1=25$

$\Rightarrow a^2+b^2=25$