Given that we need to show the points A(3, - 2), B(1,0), C(- 1, - 2) and D(1, - 4) are con - cyclic.

The term con - cyclic means the points lie on the same circle.

Let us make a circle with any three points and check whether the fourth point lies on it or not.

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

We know that the standard form of the equation of the circle is given by:

⇒ x² + y² + 2ax + 2by + c = 0 ..... (1)

Substituting A(3, - 2) in (1), we get,

⇒ 3² + (- 2)² + 2a(3) + 2b(- 2) + c = 0

⇒ 9 + 4 + 6a - 4b + c = 0

⇒ 6a - 4b + c + 13 = 0 ..... (2)

Substituting B(1,0) in (1), we get,

⇒ 1² + 0² + 2a(1) + 2b(0) + c = 0

⇒ 1 + 2a + c = 0 ......- (3)

Substituting C(- 1, - 2) in (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)

On solving (2), (3) and (4) we get,

⇒ a = - 1, b = 2 and c = 1

Substituting these values in (1), we get

⇒ x² + y² + 2(- 1)x + 2(2)y + 1 = 0

⇒ x² + y² - 2x + 4y + 1 = 0 ..... (5)

Substituting D(1, - 4) in eq(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.