Given that we need to find the equation of the circle passing through the points (0,0), (- 2,1) and (- 3, 2).

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

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

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

⇒ 0² + 0² + 2a(0) + 2b(0) + c = 0

⇒ 0 + 0 + 0a + 0b + c = 0

⇒ c = 0 ..... (2)

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

⇒ (- 2)² + 1² + 2a(- 2) + 2b(1) + c = 0

⇒ 4 + 1 - 4a + 2b + c = 0

⇒ - 4a + 2b + c + 5 = 0 ..... (3)

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

⇒ (- 3)² + 2² + 2a(- 3) + 2b(2) + c = 0

⇒ 9 + 4 - 6a + 4b + c = 0

⇒ - 6a + 4b + c + 13 = 0 ..... (4)

Solving (2), (3), (4) we get

⇒ .

Substituting these values in (1), we get

⇒

⇒ x² + y² - 3x - 11y = 0

∴ The equation of the circle is x² + y² - 3x - 11y = 0.