JEE Main 2022 — Circle Question with Solution

Note:

For equation of circle $$x^2+y^2+2gx+2fy+c=0$$, center is $$(-g,-f)$$ and radius $$r=\sqrt{g^2+f^2-c}$$

Given,

equation of circle is

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

$$\Rightarrow x^2+y^2-\sqrt{2}x-\sqrt{2}y=0$$

$$\Rightarrow x^2+y^2+2\left(-\frac{1}{\sqrt{2}}\right)x+2\left(-\frac{1}{\sqrt{2}}\right)=0$$

$$\therefore$$ $$g=-\frac{1}{\sqrt{2}}$$ and $$f=-\frac{1}{\sqrt{2}}$$

$$\therefore$$ Center $$=(-g,-f)=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$$

And Radius $$=r=\sqrt{{\left(-\frac{1}{\sqrt{2}}\right)}^2+{\left(-\frac{1}{\sqrt{2}}\right)}^2-0}$$

$$=\sqrt{\frac{1}{2}+\frac{1}{2}}=\sqrt{1}=1$$

As AB and AC makes an angle 90$${}^{\circ }$$ then line BC passes through the center of circle and BC is the diameter of the circle.

$$\therefore$$ Length of BC = 2r = 2 $$\times$$ 1 = 2

$$\therefore$$ AC² = BC² $$-$$ AB²

= 2² $$-$$ ($$\sqrt{2}$$)²

= 2

$$\Rightarrow$$ AC = $$\sqrt{2}$$

$$\therefore$$ Area of right angle triangle ABC

= $$\frac{1}{2}$$ $$\times$$ AC $$\times$$ AB

= $$\frac{1}{2}$$ $$\times$$ $$\sqrt{2}$$ $$\sqrt{2}$$

= 1 square unit.