> 
> Let $a, b, x, y$ be real numbers such $a^2+b^2=81$, $x^2+y^2=121$ and $ax + by = 99$. The set of
> all possible values of $ay - bx$ is:
> 
> 
> A. $(0,9/11]$
> 
> 
> B. $(0,9/11)$
> 
> 
> C. ${0}$
> 
> 
> D. $[9/11, \infty)$
> 
> 
> 

I have already solved the problem and obtained $0$ as the correct answer. I did it by first multiplying the first two equations and then subtracting the square of the third from it. This gave me $(ay-bx)^2=0$.

However, when I try to substitute in values such as $a=±9,b=0,x=±11,y=0$ I found the given expression to be $=±99$, and more generally values other than $0$. Although it's almost definitely something trivial, I just don't understand what am I missing here. Can someone please help me out? Thanks