Given:

The points (3, -2), (1, 0), (-1, -2) and (1, -4)

Let us assume the circle passes through the points A, B, C.

So by using the standard form of the equation of the circle:

x² + y² + 2ax + 2by + c = 0….. (1)

Substitute the points A (3, – 2) in equation (1), we get,

3² + (-2)² + 2a(3) + 2b(-2) + c = 0

9 + 4 + 6a – 4b + c = 0

6a – 4b + c + 13 = 0….. (2)

Substitute the points B (1, 0) in equation (1), we get,

1² + 0² + 2a(1) + 2b(0) + c = 0

1 + 2a + c = 0 ……- (3)

Substitute the points C (-1, -2) in equation (1), we get,

(- 1)² + (- 2)² + 2a(- 1) + 2b(- 2) + c = 0

1 + 4 – 2a – 4b + c = 0

5 – 2a – 4b + c = 0

2a + 4b – c – 5 = 0….. (4)

Upon simplifying the equations (2), (3) and (4) we get,

a = – 1, b = 2 and c = 1

Substituting the values of a, b, c in equation (1), we get

x² + y² + 2(- 1)x + 2(2)y + 1 = 0

x² + y² – 2x + 4y + 1 = 0 ….. (5)

Now by substituting the point D (1, -4) in equation (5) we get,

1² + (- 4)² – 2(1) + 4(- 4) + 1

1 + 16 – 2 – 16 + 1

0

∴ The points (3, -2), (1, 0), (-1, -2), (1, -4) are con – cyclic.