Question

Answer the following :

Show that the circles touch each other internally. Find their point of contact and the equation of their common tangent:

x² + y² – 4x – 4y – 28 = 0,

x² + y² – 4x – 12 = 0

Answer 1

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 C₁ = (2, 2)

Radius of the first circle is

r₁ = `sqrt((-2)^2 + (-2)^2 + 28)`

= `sqrt(4 + 4 + 28)`

= `sqrt(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)`

= `sqrt(16)`

= 4

By distance formula,

C₁C₂ = `sqrt((2 - 2)^2 + (0 - 2)^2`

= `sqrt(4)`

 = 2

|r₁ – r₂| = 6 – 4  = 2

Since, C₁C₂ = |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.