Let the orthocentre and centroid of a triangl | vedclass

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(B) Given that the orthocentre $A$ is $(-3, 5)$ and the centroid $B$ is $(3, 3)$.The distance $AB = \sqrt{(3 - (-3))^2 + (3 - 5)^2} = \sqrt{6^2 + (-2)

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$3\sqrt{\frac{5}{2}}$

Vedclass · Answer

(B) Given that the orthocentre $A$ is $(-3, 5)$ and the centroid $B$ is $(3, 3)$.The distance $AB = \sqrt{(3 - (-3))^2 + (3 - 5)^2} = \sqrt{6^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}$.The centroid $B$ divides the line segment joining the orthocentre $A$ and the circumcentre $C$ in the ratio $2:1$,such that $AB:BC = 2:1$.This implies $AB = \frac{2}{3}AC$,or $AC = \frac{3}{2}AB$.Substituting the value of $AB$,we get $AC = \frac{3}{2}(2\sqrt{10}) = 3\sqrt{10}$.The radius of the circle with $AC$ as diameter is $r = \frac{AC}{2} = \frac{3\sqrt{10}}{2}$.We can rewrite this as $r = 3\sqrt{\frac{10}{4}} = 3\sqrt{\frac{5}{2}}$.

Let the orthocentre and centroid of a triangle be $A(-3, 5)$ and $B(3, 3)$ respectively. If $C$ is the circumcentre of this triangle,then the radius of the circle having line segment $AC$ as diameter is:

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