My approach is as folow
Hence as per the power rule the Centriod cuts the line joining Orthocentre and Circumcentre in the ratio 2:1 hence $G=(\frac{14}{3},\frac{23}{3})$ but this lies on the line $2x+y=17$ hence I presume that it is a typo error just want to take expert advice
Perpendicular bisector of the side $BC$ is passing through the circumcenter $O(6,10)$. The line $\overleftrightarrow{BC}$ is $y=-2x+17$ so the perpendicular bisector is $y=\frac12x+7.$ The intersection of these lines is $A'=(4,9)$, the midpoint of $BC$.
On the other hand, by $HG:GO=2:1$ collinearity property, $G=\frac13H+\frac23O=(\frac{14}{3},\frac{23}{3})$ where $H=(2,3)$ is the orthocenter.
By $AG:GA'=2:1$ collinearity property, $G=\frac13A+\frac23A'$. Hence, $A=3G-2A'=(6,5).$ Now, we notice that $A\in\overleftrightarrow{BC}$ and the triangle is degenerate.
You're on the right path, Orthocentre(H) and circumcentre(O) are divided in $2:1$ ratio by the centroid(G). So in your case, lets draw a perpendicular median through A meeting BC at D, AD's slope is $\frac{1}{2}$ using the fact that $2x+y=17$ has a slope $-2$. Now circumradius is the distance between circumcentre and any vertex (here we are going to use vertex A). Solving AD and BC we get point $D(6,5)$. Let A be $(m,n)$. Now as centroid divides AD in 2:1 too, we can use the section formula to find A as (-2,6) [by comparing the centroid we got initally which is $G(\frac{10}{3}, \frac{16}{3}) ]$. To conclude find the distance between A and O and you should be getting Circumradius = 5 units.
