Presume you have x,y values for each shot from an arbitrary origin

1) calculate the straight line distance between each shot and all of the others (that's 9 + 8 + 7....) values. Probably easiest using rectangular notation by simply subtracting x from x and y from y and then converting each value into polar notation.

2) Find the three largest values of magnitude (disregard the angle)

3) these three shots form the vertices of a triangle

4) The centre of the triangle is the centre of the shot group (classical geometry)

5) Calculate the circle which will circumscribe the triangle (classical geometry)

6) correct the circle size to allow for the diameter of each shot hole. (I think in practice it is only necessary to consider the shot which is furthest from the circle centre - any circle that encloses the outside edge of this shot must also enclose the outside edge of the others.....I think)

You now have the centre and size of the smallest circle which will encompass all shots.