Let the equation of the circle passing through the points (3, -2), (1, 0) and (-1, -2) be 

x² + y² + 2gx + 2fy + c = 0 …..(i) 

For point (3, -2), 

Substituting x = 3 and y = -2 in (i), we get 9 + 4 + 6g – 4f + c = 0 

⇒ 6g – 4f + c = -13 ….(ii) 

For point (1, 0), 

Substituting x = 1 and y = 0 in (i), we get 1 + 0 + 2g + 0 + c = 0 

⇒ 2g + c = -1 ……(iii) 

For point (-1, -2), 

Substituting x = -1 and y = -2, we get 1 + 4 – 2g – 4f + c = 0

⇒ 2g + 4f – c = 5 …….(iv) 

Adding (ii) and (iv), we get 8g = -8 

⇒ g = -1 

Substituting g = -1 in (iii), we get -2 + c = -1 

⇒ c = 1 

Substituting g = -1 and c = 1 in (iv), we get -2 + 4f – 1 = 5 

⇒ 4f = 8 

⇒ f = 2 

Substituting g = -1, f = 2 and c = 1 in (i), we get 

x² + y² – 2x + 4y + 1 = 0 ……….(v) 

If (1, -4) satisfies equation (v), the four points are concyclic.

Substituting x = 1, y = -4 in L.H.S of (v), we get 

L.H.S. = (1)² + (-4)² – 2(1) + 4(-4) + 1 

= 1 + 16 – 2 – 16 + 1 

= 0 

= R.H.S. 

Point (1, -4) satisfies equation (v). 

∴ The given points are concyclic.