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

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