An inequality problem

Let $a\quad =9\cos { \alpha } \\ \\ b\quad =9\sin { \alpha } \\ \\ x\quad =11\cos { \beta } \\ \\ y\quad =11\sin { \beta }$ .

$\\ \because \quad ax\quad +\quad by\quad =\quad 99\\ \Rightarrow \quad \beta -\alpha \quad =\quad 2n\pi \quad \quad \quad (\quad n\quad \in \quad I\quad )\\ \quad \\ \Rightarrow \quad ay-bx\quad =\quad 99\sin { (\beta -\alpha ) } \\ \Rightarrow \quad \quad \quad \quad \quad \quad \quad =\quad 0\quad \quad$ .

Q.E.D

$(ax+by)^2+(ay-bx)^2=(a^2+b^2)(x^2+y^2)$

or, $(ay-bx)^2=121^2+81^2-99^2$

or, $ay-bx=0$

This problem becomes trivial by Cauchy's Inequality.: Given $a,b,x,y$ as reals. So, we know by cauchy's inequality, that $(a^2+b^2)(x^2+y^2)\geq(ax+by)^{2}$ . Where, equality holds, if $\dfrac{a}{x}=\dfrac{b}{y}$ . In this ques. we know that the equality holds, by given values. So. $\dfrac{a}{x}=\dfrac{b}{y}$ $ay=bc$ $ay-bx=0$

I assumed a as $9$ and x as $11$ and it satisfied the next equation. So according to these values the req. Answer can be 0 and only one option includes 0