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

We are given a circle $$x^2 - \sqrt{2}(x + y) + y^2 = 0$$ with an inscribed triangle $$ABC$$ where $$\angle BAC = \frac{\pi}{2}$$ and $$AB = \sqrt{2}$$.

Rewriting the circle equation in standard form, $$x^2 - \sqrt{2}x + y^2 - \sqrt{2}y = 0$$ so that $$(x - \frac{\sqrt{2}}{2})^2 + (y - \frac{\sqrt{2}}{2})^2 = \frac{1}{2} + \frac{1}{2} = 1$$ and hence the center is $$\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$$ with radius $$R = 1$$.

Since $$\angle BAC = \frac{\pi}{2}$$, the chord $$BC$$ subtends a right angle at a point on the circle and by the inscribed angle theorem must be a diameter, so $$BC = 2R = 2$$.

Triangle $$ABC$$ is right-angled at $$A$$ with $$AB = \sqrt{2}$$ and hypotenuse $$BC = 2$$, and by the Pythagorean theorem $$AC = \sqrt{BC^2 - AB^2} = \sqrt{4 - 2} = \sqrt{2}$$.

Therefore, the area is $$\frac{1}{2} \times AB \times AC = \frac{1}{2} \times \sqrt{2} \times \sqrt{2} = \frac{1}{2} \times 2 = 1$$.

Therefore, the answer is Option A: $$\textbf{1}$$.