Let a triangle ABC be inscribed in the circle x ^ 2 - 2 (x + y) + y...

Given,

circle ${x}^{2} - \sqrt{2}\left(x + y\right) + {y}^{2} = 0$

With centre \left(\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}\right) and radius $1$.

Since $∠ BAC = \dfrac{\pi}{2}$, so $BC$ will act as diameter,

$BC =$diameter$= 2$ (given)

So, $AC = \sqrt{B{C}^{2} - A{B}^{2}} = \sqrt{2}$ and $AB = \sqrt{2}$

Now, $△ ABC = \dfrac{1}{2}AB \cdot AC = 1$