Question

Answer the following :

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

x² + y² – 4x + 10y +20 = 0,

x² + y² + 8x – 6y – 24 = 0.

Answer 1

Given equation of the first circle is

x² + y² – 4x + 10y +20 = 0

Here, g = – 2, f = 5, c = 20

Centre of the first circle is C₁ = (2, – 5)

Radius of the first circle is r₁ = `sqrt((-2)^2 + 5^2 - 20)`

= `sqrt(4 + 25 - 20)`

= `sqrt(9)`

= 3

Given equation of the second circle is

x² + y² + 8x – 6y – 24 = 0

Here, g = 4, f = – 3, c = – 24

Centre of the second circle is C₂ = (–4, 3)

Radius of the second circle is

r₂ = `sqrt(4^2 + (-3)^2 + 24)`

= `sqrt(16 + 9 + 24)`

= `sqrt(49)`

= 7

By distance formula,

C₁C₂ =`sqrt((-4 - 2)^2+ [3 - (-5)]^2`

= `sqrt(36 + 64)`

= `sqrt(100)`

= 10

r₁ + r₂ = 3 + 7 = 10

Since, C₁C₂ = r₁ + r₂

∴ the given circles touch each other externally.

Let P(x, y) be the point of contact.

∴ P divides C₁ C₂ internally in the ratio r₁ : r₂ i.e. 3:7

∴ By internal division,

x = `(3(-4) + 7(2))/(3 + 7) = (-12 + 14)/10 = 1/5`

and y = `(3(3) + 7(5))/(3 + 7) = (9 - 35)/10 = -13/5`

∴ Point of contact = `(1/5, -13/5)`

Equation of common tangent is

(x² + y² – 4x + 10y + 20) – (x² + y² + 8x – 6y – 24) = 0

∴ – 4x + 10y + 20 – 8x + 6y + 24 = 0

∴ – 12x + 16y + 44 = 0

∴ 3x – 4y – 11 = 0