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.