Question

circumference of the circle touching the lines 3x+4y-3=0 , 6x+8y-1=0 is

Answer 1

Question :- find circumference of the circle touching the lines 3x+4y-3=0 , 6x+8y-1=0 is ?

Answer :-

Given Lines are :-

→ 3x + 4y - 3 = 0

→ 3x + 4y = 3

and,

→ 6x + 8y - 1 = 0

→ 6x + 8y = 1

→ 2(3x + 4y) = 1

→ (3x + 4y) = (1/2)

Therefore, both the Lines are parallel to Each other.

Now, we know that,

• Diameter of circle between parallel lines ax + by = c1 and ax + by = c2 is :- { | c1 - c2 | / √(a² + b²) }

we have :-

• a = 3
• b = 4
• c1 = 3
• c2 = (1/2)

Putting values now , we get :-

→ Diameter of circle = { | 3 - (1/2) | / (√(3² + 4²) }

→ D = { (5/2) / √(25) }

→ D = (5/2) / 5

→ D = (5/2) * (1/5)

→ D = (1/2) unit.

Therefore,

→ Radius of circle = Diameter / 2 = (1/2) * (1/2) = (1/4) unit.

Hence,

→ Circumference of circle = 2 * π * radius

→ Circumference = 2 * (22/7) * (1/4)

→ Circumference = (44 / 7 * 4)

→ Circumference = (11/7) units. (Ans.)

∴ Circumference of given circle is (11/7) units.