Central attention

Let D be the mid point of BC, therefore the centre of the circle. Angles BGC is angle in semi circle so it's a right angle. Using Pythagoras, we have the the radius of the circle $BD=DC=DG=\frac{17}{2}$ Now since the centroid divides the median in 2:1, so we can find out the median $AD=3DG=\frac{51}{2}$ Now using the famous Apollonius theorem $AB^2+AC^2=2(AD^2+BD^2)$ We have $AB^2+AC^2=2((\frac{51}{2})^2+(\frac{17}{2})^2)=\boxed{1445}$

NB: Please draw a diagram when viewing this solution, so as to increase the comprehensibility of this solution.

Let D, E, F be the mid-points of the edges opposite to A, B, C respectively.

It is obvious that AGD, BGE, CGF are straight lines.

Since G is the centroid, FG=0.5CG=7.5 and EG=0.5BG=4.

Since the circle passes through G, arcBGC is a semicircle. Therefore, angleBGC is a right angle.

Therefore, angleEGC and angleFGB are also right angles.

Therefore, triangleEGC and triangleFGB are right triangles. (yds)

Therefore, EC = sqrt(241) by the Pythagoras' Theorem.

Similarly, FB = sqrt(120.25)

Since E and F are the mid-points of the sides they are lying on respectively, the length of AB and of AC are trivial.

Therefore the answer is trivial.

Therefore everything is trivial.