Let a triangle ABC be inscribed in the circle | vedclass

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(D) The equation of the circle is $x^{2} + y^{2} - \sqrt{2}x - \sqrt{2}y = 0$.Comparing this with the standard form $x^{2} + y^{2} + 2gx + 2fy + c = 0

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(D) The equation of the circle is $x^{2} + y^{2} - \sqrt{2}x - \sqrt{2}y = 0$.Comparing this with the standard form $x^{2} + y^{2} + 2gx + 2fy + c = 0$,we get $g = -\frac{\sqrt{2}}{2} = -\frac{1}{\sqrt{2}}$ and $f = -\frac{1}{\sqrt{2}}$.The radius $r$ of the circle is $\sqrt{g^{2} + f^{2} - c} = \sqrt{(-\frac{1}{\sqrt{2}})^{2} + (-\frac{1}{\sqrt{2}})^{2} - 0} = \sqrt{\frac{1}{2} + \frac{1}{2}} = 1$.Since $\angle BAC = \frac{\pi}{2}$,the side $BC$ is the diameter of the circle.Therefore,$BC = 2r = 2(1) = 2$.In the right-angled triangle $ABC$,by the Pythagorean theorem,$AC = \sqrt{BC^{2} - AB^{2}} = \sqrt{2^{2} - (\sqrt{2})^{2}} = \sqrt{4 - 2} = \sqrt{2}$.The area of $	riangle ABC = \frac{1}{2} 	imes AB 	imes AC = \frac{1}{2} 	imes \sqrt{2} 	imes \sqrt{2} = \frac{1}{2} 	imes 2 = 1$.

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

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