Answer :

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.


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