# 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. - Mathematics and Statistics

Sum

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

Is there an error in this question or solution?

#### APPEARS IN

Balbharati Mathematics and Statistics 1 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 6 Circle
Miscellaneous Exercise 6 | Q II. (12) (ii) | Page 138