Show that the circles touch each other internally. Find their point of contact and. x^2 + y^2 + 4x – 12y + 4 = 0,

Question

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 =\(\sqrt{2^2+(-6)^2-4}\)

= \(\sqrt{4+36-4}\)

= √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 = \(\sqrt{(-1)^2+ (-2)^2-4}\)

= \(\sqrt{9+ 16}\)

= √25 

= 5 

|r₁ – r₂| = 6 – 1 = 5 

Since, C₁ C₂ = |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

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