Given that the equation of the circle is:

⇒ λx² + (2λ - 3)y² - 4x + 6y - 1 = 0 ..... (1)

Comparing with the standard equation of circle:

⇒ x² + y² + 2ax + 2by + c = 0

We get

⇒ λ = 2λ - 3

⇒ 2λ - λ = 3

⇒ λ = 3

Substituting λ value in (1), we get

⇒ 3x² + (2(3) - 3)y² - 4x + 6y - 1 = 0

⇒ 3x² + 3y² - 4x + 6y - 1 = 0

⇒

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

⇒ Centre = (- a, - b)

⇒ Radius =

⇒ Centre(C) =

⇒

∴The correct option is (b)