The orthocentre and centroid of a triangle ar | vedclass

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(C) The circumcentre $C$ divides the line segment joining the orthocentre $A(-3, 5)$ and the centroid $B(3, 3)$ in the ratio $2:1$ externally,such tha

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$2 \sqrt{10}$

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(C) The circumcentre $C$ divides the line segment joining the orthocentre $A(-3, 5)$ and the centroid $B(3, 3)$ in the ratio $2:1$ externally,such that $B$ lies between $A$ and $C$. Using the section formula for external division,the coordinates of $C(x, y)$ are given by: $x = \frac{2(3) - 1(-3)}{2 - 1} = \frac{6 + 3}{1} = 9$ $y = \frac{2(3) - 1(5)}{2 - 1} = \frac{6 - 5}{1} = 1$ So,$C = (9, 1)$. The diameter of the circle is the length of the segment $AC$. $AC = \sqrt{(9 - (-3))^2 + (1 - 5)^2} = \sqrt{12^2 + (-4)^2} = \sqrt{144 + 16} = \sqrt{160} = 4\sqrt{10}$. The radius of the circle is $\frac{1}{2} AC = \frac{1}{2} 	imes 4\sqrt{10} = 2\sqrt{10}$.

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

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