Find the equation of family of circles which are passing through the intersection of two circles S1: x2 + y2 - 4 = 0 and S2: x2 + y2 - 2x - 4y + 4 = 0 ?

CONCEPT:

The general second degree equation in x and y, ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a circle with centre (-g, -f) and radius \(r = \sqrt {{g^2} + {f^2} - c} \), when a = b and h = 0.

The equation of the family of circles passing through the intersection of two circles S1 and S2 is given by: S1 + λS2 = 0 where λ ≠ 1.

Note: If λ = 1, the equation S1 + λS2 will represent the equation of the familiar chord of the two circles.

CALCULATION:

Given:

S1: x2 + y2 - 4 = 0 and S2: x2 + y2 - 2x - 4y + 4 = 0 are two circles.

Here, we have to find the equation of the circle which passes through the intersection of the given circles S1 and S2 

As we know, the equation of the family of circles passing through the intersection of two circles S1 and S2 is given by:

S1 + λS2 = 0 where λ ≠ 1.

Let's find out S1 + λS2 = 0

⇒ (x2 + y2 - 4) + λ(x2 + y2 - 2x - 4y + 4) = 0

⇒ (1 + λ) x2 + (1 + λ)y2 - 2λx - 4λy + (4λ - 4) = 0

So, the equation of the family of circles are: 

(1 + λ) x2 + (1 + λ) y2 - 2λx - 4λy + (4λ - 4) = 0

Hence, option D is the correct answer.