Mathematics - Circle Question with Solution | TestHub

Given,

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

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

Since $\angle BAC=\frac{\pi }{2}$, 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, $△ABC=\frac{1}{2}AB·AC=1$