If point A and B are $(1, 0)$ and B $(0, 1)$ . If point C is on the circle $x^{2} + y^{2} = 1$ , then locus of the orthocentre of the triangle ABC is

x^{2} + y^{2} = 4

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x^{2} + y^{2} - x - y = 0

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x^{2} + y^{2} - 2 x - 2 y + 1 = 0

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x^{2} + y^{2} + 2 x - 2 y + 1 = 0

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Solution

The correct option is C $x^{2} + y^{2} - 2 x - 2 y + 1 = 0$
Since point C passes through the circle we can assume C as $(cos θ, sin θ)$ where as H as $(h, k)$ which is the orthocentre of the $Δ A B C$ .
Since circumcentre of the triangle is $(0, 0)$ , for orthocentre $h = 1 + cos θ$ and $k = 1 + sin θ$ .
Eliminating $θ$ and adding the above two
$(x - 1)^{2} + (y - 1)^{2} = 1$
$∴ x^{2} + y^{2} - 2 x - 2 y + 1 = 0$

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If point A and B are (1,0) and B(0,1). If point C is on the circle x2+y2=1, then locus of the orthocentre of the triangle ABC is

If point A and B are $(1, 0)$ and B $(0, 1)$ . If point C is on the circle $x^{2} + y^{2} = 1$ , then locus of the orthocentre of the triangle ABC is