JEE Main 29 June 2022 Shift 2 Solved Paper

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
=‌

1

2

×AC×AB
=‌

1

2

×√2√2
=1‌ square unit. ‌