Given equation of the first circle is x² + y² – 4x – 4y – 28 = 0 

Here, g = -2, f = -2, c = -28 

Centre of the first circle is C = (2, 2) 

Radius of the first circle is 

r₁ = \(\sqrt{(-2)^2 + (-2)^2+28}\)

\(\sqrt{4+ 4+28}\)

= √36 

= 6 

Given equation of the second circle is x² + y² – 4x – 12 = 0

Here, g = -2, f = 0, c = -12 

Centre of the second circle is C₂ = (2, 0) 

Radius of the second circle is

r₂ = \(\sqrt {(-2)^2 + 0^2 + 12}\)

= \(\sqrt {4+ 12}\)

= 4 By distance formula,

C₁ C₁ = \(\sqrt{(2-2)^2 + (0-2)^2}\)

= √4 

= 2

r₁ – r₂ | = 6 – 4 = 2 

Since, C₁ C₂ = |r₁ – r₂ | 

∴ the given circles touch each other internally. 

Equation of common tangent is 

(x² + y² – 4x – 4y – 28) – (x² + y² – 4x – 12) = 0 

⇒ -4x – 4y – 28 + 4x + 12 = 0 

⇒ -4y – 16 = 0 

⇒ y + 4 = 0 

⇒ y = -4 

Substituting y = -4 in x² + y² – 4x – 12 = 0, we get 

⇒ x² + (-4)² – 4x – 12 = 0 

⇒ x² + 16 – 4x – 12 = 0 

⇒ x² – 4x + 4 = 0 . 

⇒ (x – 2)² = 0 

⇒ x = 2 

∴ Point of contact is (2, -4) and equation of common tangent is y + 4 = 0.