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.
Solution
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
