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 + 19 = 0,

x² + y² + 2x + 8y – 23 = 0.

Answer 1

Given equation of the first circle is

x² + y² – 4x – 10y + 19 = 0

Here, g = – 2, f = – 5, c = 19

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

Radius of the first circle is

r₁ = `sqrt((-2)^2 + (-5)^2 - 19)`

= `sqrt(4 + 25 - 19)`

= `sqrt(10)`.

Given equation of the second circle is

x² + y² + 2x + 8y – 23 = 0

Here, g = 1, f = 4, c = – 23

Centre of the second circle is C₂ = (-1, -4)

Radius of the second circle is

r₂ = `sqrt((-1)^2 + 4^2 + 23)`

= `sqrt(9 + 81)`

= `sqrt(40)`

= `2sqrt(10)`

By distance formula,

C₁C₂ = `sqrt((-1 - 2)^2 + (-4 - 5)^2`

= `sqrt(9 + 81)`

= `sqrt(90)`

= `3sqrt(10)`

r₁ + r₂ = `sqrt(10) + 2sqrt(10)`

= `3sqrt(10)`

Since, C₁C₂ = r₁ + r₂

∴ the given circles touch each other externally.

r₁ : r₂ = `sqrt(10) : 2sqrt(10)` = 1 : 2

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

∴ P divides C₁ C₂ internally in the ratio r₁ : r₂ i.e. 1:2

∴ By internal division,

x = `(1(-1) + 2(2))/(1 + 2) = (-1 + 4)/3` = 1

an y = `(1(-4) + 2(5))/(1 + 2) = (-4 + 10)/3` = 2

∴ Point of contact = (1, 2)

Equation of common tangent is

(x² + y² – 4x – 10y + 19) – (x² + y² + 2x + 8y – 23) = 0

∴ – 4x – 10y + 19 – 2x – 8y + 23 = 0

∴ – 6x – 18y + 42 = 0

∴ x + 3y – 7 = 0