The distance between the circumcentre and the centroid of the triangle formed by the vertices (1,2), (3, - 1) and (4,0), is

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$\frac{11 \sqrt{2}}{30}$

$\frac{9 \sqrt{2}}{5}$

$\frac{1}{\sqrt{2}} \sqrt{45}$

answer is C.

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Detailed Solution

Let A=(1 ,2),B= (3,- 1), C =(4,,0) = Centroid $G = (\frac{8}{3}, \frac{1}{3})$
S=(α,ß) be the circumcentre of $△ ABC \Rightarrow SA = SB = SC$
$\begin{array}{l} \Rightarrow (α - 1)^{2} + (β - 2)^{2} = (α - 3)^{2} + (β + 1)^{2} = (α - 4)^{2} + β^{2} \\ \Rightarrow - 2 α - 4 β + 5 = - 6 α + 2 β + 10 = - 8 α + 16 \Rightarrow - 4 α - 6 β - 5 = 0, 2 α + 2 β - 6 = 0 \\ \Rightarrow α = \frac{23}{10}, β = \frac{7}{10} . ∴ S = (\frac{23}{10}, \frac{7}{10}) \end{array}$
Now $GS = \sqrt{{(\frac{8}{3} - \frac{23}{10})}^{2} + {(\frac{1}{3} - \frac{7}{10})}^{2}} = \sqrt{{(\frac{80 - 69}{30})}^{2} + {(\frac{10 - 21}{30})}^{2}} = \frac{1}{30} \sqrt{121 + 121} = \frac{11 \sqrt{2}}{30} .$