(a) First find the equation of the circle passing through the points 

(0,0) (1, 1) (5,-5) (6, -4) 

(0,0) ⇒ C = 0 …. (1) 

(1, 1) ⇒ 2g + 2f + 2 = 0 …. (2) 

(5,-5) ⇒ 10g -10f + 50 + c = 0 …. (3)

∴ The equation of the circle is x² + y² – 6x + 4y = 0 

Substituting the fourth point (6,-4) we get 

36 + 16 – 36 – 16 = 0 ⇒ 0 = 0 

∴ The points are concyclic. 

(b) Let us find the equation of the circle passing through (2,-4) (3,- 1) and (3, -3) we get 

(2,-4)4g – 8f + C + 20 = 0 … (1) 

(3,-1) 6g – 2f+ C + 10 = 0) … (2) 

(3,-3) 6g – 6f+c= -18 … (3) 

Solving the above equations we get g = -1, f = 2 and C = 0 

∴ Required circle is x² + y² – 2x + 4y = 0 

Substitute the fourth point (0, 0), we get 0 + 0 + 0 + 0 = 0 

∴ The four points are concyclic 

(c) Let us find the equation of the circle passing through (1, 0) (2, -7) and (8, 1) we get 

(1,0) 2g + C + 2 = 0 … (1) 

(2,-7) 4g – 14f + C + 53 = 0 … (2) 

(8,1) 16g + 2F + C + 25 = 0 … (3) 

Solving above 3 equations we get g = -\(\frac{738}{50}\),f = \(\frac{147}{50}\)and c = \(\frac{196}{25}\)

∴ Required circle is x² + y² + \(\frac{246}{50}\)x + \(\frac{147}{25}\)y + \(\frac{196}{25}\)= 0 

25x² + 25y² + 246x + 147y + 196 = 0 

Substitute the fourth point (9,-6) we get 

25(9)² + 25(-6)² + 246(9) + 147(-6) + 196 = 0 

2025 + 900 + 2214 – 882 + 196 = 0 

The points are not concyclic