show that the 4 points (1,0),(9,-6),(2,-7) and (8,1) are coincyclic points

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

Now 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 and now 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 =8^{2}+1^{2}-10*8+6*1+9

= 0

And we find it satisfy the eqn of circle. Thus all points are coincyclic.

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