# Answer the following : Show that the circles touch each other internally. Find their point of contact and the equation of their common tangent: x2 + y2 – 4x – 4y – 28 = 0, x2 + y2 – 4x – 12 = 0 - Mathematics and Statistics

Sum

Show that the circles touch each other internally. Find their point of contact and the equation of their common tangent:

x2 + y2 – 4x – 4y – 28 = 0,

x2 + y2 – 4x – 12 = 0

#### Solution

Given equation of the first circle is

x2 + y2 – 4x – 4y – 28 = 0

Here, g = – 2, f = – 2, c = – 28

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

Radius of the first circle is

r1 = sqrt((-2)^2 + (-2)^2 + 28)

= sqrt(4 + 4 + 28)

= sqrt(36)

= 6

Given equation of the second circle is

x2 + y2 – 4x – 12 = 0

Here, g = – 2, f = 0, c = – 12

Centre of the second circle is C2 = (2, 0)

Radius of the second circle is

r2 = sqrt((-2)^2 + 0^2 + 12)

= sqrt(4 + 12)

= sqrt(16)

= 4

By distance formula,

C1C2 = sqrt((2 - 2)^2 + (0 - 2)^2

= sqrt(4)

= 2

|r1 – r2| = 6 – 4  = 2

Since, C1C2 = |r1 – r2

∴ the given circles touch each other internally.

Equation of common tangent is

(x2 + y2 – 4x – 4y – 28) – (x2 + y2 – 4x – 12) = 0

∴ – 4x – 4y – 28 + 4x + 12 = 0

∴ – 4y – 16 = 0

∴ y + 4 = 0

∴ y = – 4

Substituting y = – 4 in x2 + y2 – 4x – 12 = 0, we get

∴ x2 + (– 4)2 – 4x – 12 = 0

∴ x2 + 16 – 4x – 12 = 0

∴ x2 – 4x + 4 = 0

∴ (x – 2)2 = 0

∴ x = 2

∴ Point of contact is (2, – 4) and equation of common tangent is y + 4 = 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. (13) (i) | Page 138
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