Find the equation of the circle which passes through (3, - 2), (- 2, 0) and has its centre on the line 2x – y = 3.

Given that we need to find the equation of the circle which passes through (3, - 2), (- 2,0) and has its centre on the line 2x - y = 3. ..... (1)

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

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

Substituting centre (- a, - b) in (1) we get,

⇒ 2(- a) - (- b) = 3

⇒ - 2a + b = 3

⇒ 2a - b + 3 = 0 ......(3)

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

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

⇒ 9 + 4 + 6a - 4b + c = 0

⇒ 6a - 4b + c + 13 = 0 ..... (4)

Substituting (- 2,0) in (2), we get

⇒ (- 2)² + 0² + 2a(- 2) + 2b(0) + c = 0

⇒ 4 + 0 - 4a + c = 0

⇒ 4a - c - 4 = 0 ..... (5)

Solving (3), (4) and (5) we get,

⇒

Substituting these values in (2), we get

⇒

⇒ x² + y² + 3x + 12y + 2 = 0

∴ The equation of the circle is x² + y² + 3x + 12y + 2 = 0.