y-x=0 is the equation of a side of a triangle | vedclass

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(D) Let the orthocentre be $H = (5, 8)$ and the circumcentre be $O = (2, 3)$.The radius $R$ of the circumcircle is the distance between the circumcent

Vedclass · Accepted Answer

$2 \sqrt{10}$

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(D) Let the orthocentre be $H = (5, 8)$ and the circumcentre be $O = (2, 3)$.The radius $R$ of the circumcircle is the distance between the circumcentre $O$ and any vertex of the triangle.We know that the reflection of the orthocentre $H$ with respect to any side of the triangle lies on the circumcircle.Let the side be $L: x - y = 0$.The reflection $(x', y')$ of $H(5, 8)$ with respect to $x - y = 0$ is given by $\frac{x' - 5}{1} = \frac{y' - 8}{-1} = -2 \frac{5 - 8}{1^2 + (-1)^2} = -2 \frac{-3}{2} = 3$.So,$x' = 5 + 3 = 8$ and $y' = 8 - 3 = 5$.The point $(8, 5)$ lies on the circumcircle.The radius $R$ is the distance between the circumcentre $O(2, 3)$ and the point $(8, 5)$ on the circle.$R = \sqrt{(8 - 2)^2 + (5 - 3)^2} = \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40} = 2 \sqrt{10}$.

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