The circumcentre of a triangle lies at the or | vedclass

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(D) Let the circumcentre be $O = (0, 0)$.The centroid $G$ is the midpoint of the segment joining $(a^2 + 1, a^2 + 1)$ and $(2a, -2a)$.$G = \left( \fra

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$(a - 1)^2x - (a + 1)^2y = 0$

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(D) Let the circumcentre be $O = (0, 0)$.The centroid $G$ is the midpoint of the segment joining $(a^2 + 1, a^2 + 1)$ and $(2a, -2a)$.$G = \left( \frac{a^2 + 1 + 2a}{2}, \frac{a^2 + 1 - 2a}{2} \right) = \left( \frac{(a + 1)^2}{2}, \frac{(a - 1)^2}{2} \right)$.We know that the orthocentre $H$,centroid $G$,and circumcentre $O$ are collinear,and $G$ divides $OH$ in the ratio $1:2$ internally.Using the section formula,$G = \frac{1 \cdot H + 2 \cdot O}{1 + 2} = \frac{H}{3}$.Thus,$H = 3G = \left( \frac{3(a + 1)^2}{2}, \frac{3(a - 1)^2}{2} \right)$.Let the line be $Ax + By = 0$. Substituting the coordinates of $H$ into the equation in option $(d)$:$(a - 1)^2 \left( \frac{3(a + 1)^2}{2} \right) - (a + 1)^2 \left( \frac{3(a - 1)^2}{2} \right) = \frac{3}{2} (a - 1)^2 (a + 1)^2 - \frac{3}{2} (a + 1)^2 (a - 1)^2 = 0$.Since the coordinates of $H$ satisfy the equation in option $(d)$,the orthocentre lies on this line.

The circumcentre of a triangle lies at the origin and its centroid is the midpoint of the line segment joining the points $(a^2 + 1, a^2 + 1)$ and $(2a, -2a)$,where $a \ne 0$. Then for any $a$,the orthocentre of this triangle lies on the line:

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