Let a triangle ABC be inscribed in the circle x^2-√2(x+y)+y^2=0 such that BAC=2.

Question

Let a triangle $ABC$ be inscribed in the circle $x^2-\sqrt2(x+y)+y^2=0$ such that $\angle BAC=\dfrac\pi2$. If the length of side $AB$ is $\sqrt2$, then the area of $	riangle ABC$ is equal to:

Accepted Answer

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

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

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

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

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

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

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

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

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 - ($$\sqrt2$$)2

= 2

$$\Rightarrow$$ AC = $$\sqrt2$$

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

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

= $${1 \over 2}$$ $$\times$$ $$\sqrt2$$ $$\sqrt2$$

= 1 square unit.

Solutions · 1