Let the equation of circle passing through the points (9, 1), (7, 9), (-2, 12) be 

x² + y² + 2gx + 2fy + c = 0 …….(i) 

For point (9, 1), 

Substituting x = 9 and y = 1 in (i), we get 

81 + 1 + 18g + 2f + c = 0 

⇒ 18g + 2f + c = -82 …..(ii) 

For point (7, 9), 

Substituting x = 7 and y = 9 in (i), we get 

49 + 81 + 14g + 18f + c = 0 

⇒ 14g + 18f + c = -130 ……(iii) 

For point (-2, 12), 

Substituting x = -2 and y = 12 in (i), we get 

4 + 144 – 4g + 24f + c = 0

⇒ -4g + 24f + c = -148 …..(iv) 

By (ii) – (iii), we get 4g – 16f = 48 

⇒ g – 4f = 12 …..(v) 

By (iii) – (iv), we get 18g – 6f = 18 

⇒ 3g – f = 3 ……(vi) 

By 3 × (v) – (vi), we get -11f = 33 

⇒ f = -3 

Substituting f = -3 in (vi), we get 3g – (-3) = 3 

⇒ 3g + 3 = 3 

⇒ g = 0 

Substituting g = 0 and f = -3 in (ii), we get 

18(0) + 2(-3) + c = – 82 

⇒ -6 + c = -82

 ⇒ c = -76 

Equation of the circle becomes 

x² + y² + 2(0)x + 2(-3)y + (-76) = 0 

⇒ x² + y² – 6y – 76 = 0 ……(vii) 

Now for the point (6, 10), 

Substituting x = 6 and y = 10 in L.H.S. of (vii),

we get L.H.S = 6² + 10² – 6(10) – 76 

= 36 + 100 – 60 – 76 

= 0 

= R.H.S. 

∴ Point (6,10) satisfies equation (vii). ∴ the given points are concyciic.