Consider a triangle whose two sides lie on the x -axis and the line x + y + 1 = 0 . If the orthocenter of is (1, 1), then the equation of the circle passing through the vertices of the triangle is [2021]

Question

Consider a triangle $Δ$ whose two sides lie on the $x$ -axis and the line $x + y + 1 = 0$ . If the orthocenter of $Δ$ is (1, 1), then the equation of the circle passing through the vertices of the triangle $Δ$ is [2021]

Smart Achievers · Accepted Answer

(2)

$(1, - 2) = (α, - α - 1) \Rightarrow α = 1$

It is clear from question that one of the vertex of triangle is intersection of $x$ -axis and $x + y + 1 = 0 \Rightarrow A (- 1, 0)$

$Let vertex B be (α, - α - 1)$

$Line A C ⊥ B H, so m_{A C} \cdot m_{B H} = - 1$

$\Rightarrow 0 = - \frac{(1 - α)}{α + 2} \Rightarrow α = 1 \Rightarrow B (1, - 2)$

$Let vertex C be (β, 0)$

$Line A H ⊥ B C$

$∴$ $m_{A H} \cdot m_{B C} = - 1$

$\Rightarrow \frac{1}{2} \cdot \frac{2}{β - 1} = - 1 \Rightarrow β = 0$

$Centroid of ∆ A B C is (0, - \frac{2}{3})$

$We know that G (centroid) divides line joining circumcentre (O) and orthocentre (H) in the ratio 1 : 2$

$\Rightarrow 2 h + 1 = 0$ $\Rightarrow \frac{2 k + 1}{3} = - \frac{2}{3}$

$\Rightarrow h = - \frac{1}{2}$ $\Rightarrow k = - \frac{3}{2}$ $\Rightarrow Circumcentre is (- \frac{1}{2}, - \frac{3}{2})$

$Equation of circumcircle is (passing hrough C (0, 0)) is$ $x^{2} + y^{2} + x + 3 y = 0$

Solution

Q.
Consider a triangle $Δ$ whose two sides lie on the $x$ -axis and the line $x + y + 1 = 0$ . If the orthocenter of $Δ$ is (1, 1), then the equation of the circle passing through the vertices of the triangle $Δ$ is [2021]

1 $x^{2} + y^{2} - 3 x + y = 0$

2 $x^{2} + y^{2} + x + 3 y = 0$

3 $x^{2} + y^{2} + 2 y - 1 = 0$

4 $x^{2} + y^{2} + x + y = 0$

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Solution

Q. Consider a triangle Δ whose two sides lie on the x-axis and the line x+y+1=0. If the orthocenter of Δ is (1, 1), then the equation of the circle passing through the vertices of the triangle Δ is [2021]

1 x2+y2-3x+y=0

2 x2+y2+x+3y=0

3 x2+y2+2y-1=0