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 – 12y + 4 = 0,

x² + y² – 2x – 4y + 4 = 0

Answer 1

Given equation of the first circle is

x² + y² + 4x – 12y + 4 = 0

Here, g = 2, f = – 6, c = 4

Centre of the first circle is C₁ = (– 2, 6)

Radius of the first circle is

r₁ = `sqrt(2^2 + (-6)^2 - 4)`

= `sqrt(4 + 36 - 4)`

= `sqrt(36)`

= 6

Given equation of the second circle is

x² + y² – 2x – 4y + 4 = 0

Here, g = – 1, f = – 2, c = 4

Centre of the second circle is C₂ = (1, 2)

Radius of the second circle is

r₂ = `sqrt((-1)^2 + (-2)^2 - 4)`

= `sqrt(1 + 4 - 4)`

= `sqrt(1)`

= 1

By distance formula

C₁C₂ = `sqrt([1 - (-2)]^2  (2 - 6)^2`

= `sqrt(9 + 16)`

= `sqrt(25)`

= 5

|r₁ – r₂| = 6 – 1 = 5

Since, C₁C₂ = |r₁ – r₂|

∴ the given circles touch each other internally.

Equation of common tangent is

(x² + y² + 4x – 12y + 4) – (x² + y² – 2x – 4y + 4) = 0

∴ 4x – 12y + 4 + 2x + 4y – 4 = 0

∴ 6x – 8y = 0

∴ 3x – 4y = 0

∴ y = `(3x)/4`

Substituting y = `(3x)/4` in x² + y² – 2x – 4y + 4 = 0, we get

`x^2 + ((3x)/4)^2 - 2x - 4((3x)/4) + 4` = 0

∴ `x^2 + (9x^2)/16 - 2x - 3x + 4` = 0

∴ `(25x^2)/16 - 5x + 4` = 0

∴ 25x² – 80x + 64 = 0

∴ (5x – 8)² = 0

∴ 5x – 8 = 0

∴ x = `8/5`

Substituting x = `8/5` in y = `(3x)/4`, we get

y = `3/4(8/5) = 6/5`

∴ Point of contact is `(8/5, 6/5)` and equation of common tangent is 3x – 4y = 0.