Find the equation of the circle which circumscribes the triangle formed by the lines:

y = x + 2, 3y = 4x and 2y = 3x

Given that we need to find the equation of the circle formed by the lines:

⇒ y = x + 2

⇒ 3y = 4x

⇒ 2y = 3x

On solving these lines we get the intersection points A(6,8), B(0,0), C(4,6)

We know that the standard form of the equation of a circle is given by:

⇒ x² + y² + 2ax + 2by + c = 0 .....(1)

Substituting (6,8) in (1), we get

⇒ 6² + 8² + 2a(6) + 2b(8) + c = 0

⇒ 36 + 64 + 12a + 16b + c = 0

⇒ 12a + 16b + c + 100 = 0 ......(2)

Substituting (0,0) in (1), we get

⇒ 0² + 0² + 2a(0) + 2b(0) + c = 0

⇒ 0 + 0 + 0a + 0b + c = 0

⇒ c = 0 .....(3)

Substituting (4,6) in (1), we get

⇒ 4² + 6² + 2a(4) + 2b(6) + c = 0

⇒ 16 + 36 + 8a + 12b + c = 0

⇒ 8a + 12b + c + 52 = 0 ..... (4)

Solving (2), (3), (4) we get

⇒ a = - 23, b = 11,c = 0.

Substituting these values in (1), we get

⇒ x² + y² + 2(- 23)x + 2(11)y + 0 = 0

⇒ x² + y² - 46x + 22y = 0

∴ The equation of the circle is x² + y² - 46x + 22y = 0.