If in a triangle \( ABC \), \( AB=5 \) units, \( \angle B=\cos^{-1}\!\left(\dfrac{3}{5}\right) \) and radius of circumcircle of \( \triangle ABC \) is \( 5 \) units, then the area (in sq. units) of \( \triangle ABC \) is:
\( 10+6\sqrt{2} \)
\( 4+2\sqrt{3} \)
\( 8+2\sqrt{2} \)
\( 6+8\sqrt{3} \)