For equation of circle x2+y2+2gx+2fy+c=0, center is (−g,−f) and radius r=√g2+f2−c Given, equation of circle is x2−√2(x+y)+y2=0 ⇒x2+y2−√2x−√2y=0 ⇒x2+y2+2(−
1
√2
)x+2(−
1
√2
)=0 ∴g=−
1
√2
and f=−
1
√2
∴ Center =(−g,−f)=(
1
√2
,
1
√2
) And Radius =r=√(−
1
√2
)2+(−
1
√2
)2−0 =√
1
2
+
1
2
=√1=1
As AB and AC makes an angle 90∘ then line BC passes through the center of circle and BC is the diameter of the circle. ∴ Length of BC=2r=2×1=2 ∴AC2=BC2−AB2 =22−(√2)2 =2 ⇒AC=√2 ∴ Area of right angle triangle ABC =
