Question

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= \(\sqrt {(-2)^2 + (-5)^2-19}\)

= \(\sqrt {4 + 25-19}\)

= √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 C = (-1, -4) 

Radius of the second circle is

r² = \(\sqrt{(-1)^2 + 4^2 + 23}\)

= \(\sqrt{1 + 16 + 23}\)

= √40 = 2√10 By distance formula,

C₁ C₂ =\(\sqrt{(-1-2)^2 + (-4-5)^2}\)

=\(\sqrt{9 + 81}\)

= √90 

= 3√10 

r₁ + r₂ = √10 + 2√10 = 3√10 

Since, C₁ C₂ = r₁ + r₂ 

the given circles touch each other externally. 

r₁ : r₂ = √10 : 2√10 = 1 : 2 

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

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