JEE Math Practice Question

Let a triangle ABC be inscribed in the circle $${x^2} - \sqrt 2 (x + y) + {y^2} = 0$$ such that $$\angle BAC = {\pi \over 2}$$. If the length of side AB is $$\sqrt 2 $$, then the area of the $$\Delta$$ABC is equal to :

1. A.1
2. B.$$\left( {\sqrt 6 + \sqrt 3 } \right)/2$$
3. C.$$\left( {3 + \sqrt 3 } \right)/4$$
4. D.$$\left( {\sqrt 6 + 2\sqrt 3 } \right)/4$$