$F_4\subseteq F_2$ and $F_4\subseteq F_3$, and $F_2\subseteq F_1,\ F_3\subseteq F_1$.
Thus $F_1\cup F_2\cup F_3\cup F_4=F_1$ (since the others are subsets of $F_1$).
For a triangle inscribed in a circle, maximum area occurs when the triangle is right-angled,
since the hypotenuse = diameter = $2r$.
Area $= \dfrac{1}{2} \times AB \times BC = \dfrac{1}{2} r \times r = r^2$.
$\boxed{\text{Answer: (B) Right angled triangle with side } 2r, r}$
$AB^{2}=14^{2}+2^{2}=200,\ BC^{2}=(-8)^{2}+6^{2}=100,\ CA^{2}=(-6)^{2}+(-8)^{2}=100$.
Since $BC=CA$ the triangle is isosceles, and $BC^{2}+CA^{2}=AB^{2}$ ⇒ right-angled at $C$.