We can prove this by using standard equation of circle.

The standard equation of the of a circle x^{2}+y^{2}+dx+ey+f=0

Put points (1,0) ,(2,-7) ,(9,-6) in equations of circle, we get three equations respectively.

1+d+f=0 ,

2d-7e+f=-53 ,

9d-6e+f=-117

Now after solving these equations, we get d= -10, e= 6, f=9.

For the proof of the concyclicity, we have to form the standard equation and then we have to check whether the fourth point satisfy the standard equation or not if it will satisfy the equation then these points are concyclic otherwise not.

Thus eqn. of circle is x^{2}+y^{2}-10x+6y+9=0

Put (8,1) in this equation =82+12-10*8+6*1+9

= 0

Thus all points are coincyclic.