Given that the equation of the circle is:

⇒ 3x² + 3y² + λxy + 9x + (λ - 6)y + 3 = 0 ...... (1)

Comparing with the standard equation of circle:

⇒ x² + y² + 2ax + 2by + c = 0

We get

⇒ λ = 0

Substituting λ value in (1), we get

⇒ 3x² + 3y² + 0xy + 9x + (0 - 6)y + 3 = 0

⇒ 3x² + 3y² + 9x - 6y + 3 = 0

⇒ x² + y² + 3x - 2y + 1 = 0

We know that for a circle x² + y² + 2ax + 2by + c = 0

⇒ Centre = (- a, - b)

⇒ Radius =

⇒ Radius (r) =

⇒

⇒

⇒

∴The correct option is (a)