My Problem Number Six

Substitute $y = 3a - x$ into $x^{2} + y^{2} = 4a^{2}$ to find that

$x^{2} + (3a - x)^{2} = 4a^{2} \Longrightarrow x^{2} + 9a^{2} - 6ax + x^{2} = 4a^{2}$

$\Longrightarrow 2x^{2} - 6ax + 5a^{2} = 0 \Longrightarrow x = \dfrac{6a \pm \sqrt{36a^{2} - 40a^{2}}}{4} = \dfrac{3a \pm a*i}{2}$ ,

which implies that $y = \dfrac{3a \mp a*i}{2}$ .

Since $a \gt 0$ we can conclude that both $x$ and $y$ must be elements of $S$ .

Let $x=2a\cos \theta$ and $y=2a\sin \theta$ . This assumption satisfies the first equation.

Plugging these values into the second equation and cancelling $a$ from both sides (since it is a positive real number), we get,

$\sin \theta + \cos \theta = \dfrac{3}{2} \\ \implies (\sin \theta + \cos \theta)^2=\dfrac{9}{4}\\ \implies 1+\sin 2\theta = \dfrac{9}{4}\\ \implies \sin 2\theta = \dfrac{5}{4} \gt 1\\ \implies \theta \notin \mathbb{R}\\ \implies \sin \theta,\cos \theta \notin \mathbb{R}\\ \implies 2a\sin \theta~,~2a\cos \theta \notin \mathbb{R}\quad [\textrm{since }2a\textrm{ is real}]\\ \implies x,y \notin \mathbb{R} \implies \boxed{x,y\in \mathbb{S}}$