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.


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