The vertices of a triangle are \((6,0),(0,6)\) and \((6,6)\). The distance between its circumcentre and centroid is

Let the vertices of a triangle be \(A(6,0), B(0,6)\) and \(C(6,6)\).
\[
\begin{array}{ll}
\text { Now, } & \mathrm{AB}=\sqrt{6^{2}+6^{2}}=6 \sqrt{2} \\
& \mathrm{BC}=\sqrt{6^{2}+0}=6 \\
\text { and } & \mathrm{CA}=\sqrt{0+6^{2}}=6 \\
\text { Also, } & \mathrm{AB}^{2}=\mathrm{BC}^{2}+\mathrm{CA}^{2}
\end{array}
\]
Therefore, \(\triangle \mathrm{ABC}\) is right angled at \(\mathrm{C}\). So, mid point of \(A B\) is the circumcentre of \(\triangle A B C\).
\(\therefore\) Coordinate of circumcentre are \((3,3)\).
Coordinate of centroid are,
\(G\left(\frac{6+0+6}{3}, \frac{0+6+6}{3}\right)\), ie, \((4,4)\)
\(\therefore\) Required distance \(=\sqrt{(4-3)^{2}+(4-3)^{2}}=\sqrt{2}\)