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

x + y + 3 = 0, x - y + 1 = 0 and x = 3

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

⇒ x + y + 3 = 0

⇒ x - y + 1 = 0

⇒ x = 3

On solving these lines we get the intersection points A(- 2, - 1), B(3,4), C(3, - 6)

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

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

Substituting (- 2, - 1) in (1), we get

⇒ (- 2)² + (- 1)² + 2a(- 2) + 2b(- 1) + c = 0

⇒ 4 + 1 - 4a - 2b + c = 0

⇒ 5 - 4a - 2b + c = 0

⇒ 4a + 2b - c - 5 = 0 ..... (2)

Substituting (3,4) in (1), we get

⇒ 3² + 4² + 2a(3) + 2b(4) + c = 0

⇒ 9 + 16 + 6a + 8b + c = 0

⇒ 6a + 8b + c + 25 = 0 ..... (3)

Substituting (3, - 6) in (1), we get

⇒ 3² + (- 6)² + 2a(3) + 2b(- 6) + c = 0

⇒ 9 + 36 + 6a - 12b + c = 0

⇒ 6a - 12b + c + 45 = 0 ..... (4)

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

⇒ a = - 3, b = 1,c = - 15.

Substituting these values in (1), we get

⇒ x² + y² + 2(- 3)x + 2(1)y - 15 = 0

⇒ x² + y² - 6x + 2y - 15 = 0

∴ The equation of the circle is x² + y² - 6x + 2y - 15 = 0.