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