Find the equation of the circle passing through the points :

(5, - 8), (- 2, 9) and (2, 1)


Given that we need to find the equation of the circle passing through the points (5, - 8), (- 2,9) and (2,1).



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


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


Substituting (5, - 8) in (1), we get


52 + (- 8)2 + 2a(5) + 2b(- 8) + c = 0


25 + 64 + 10a - 16b + c = 0


10a - 16b + c + 89 = 0 ..... (2)


Substituting (- 2,9) in (1), we get


(- 2)2 + 92 + 2a(- 2) + 2b(9) + c = 0


4 + 81 - 4a + 18b + c = 0


- 4a + 18b + c + 85 = 0 ..... (3)


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


22 + 12 + 2a(2) + 2b(1) + c = 0


4 + 1 + 4a + 2b + c = 0


4a + 2b + c + 5 = 0 ..... (4)


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


a = 58, b = 24,c = - 285.


Substituting these values in (1), we get


x2 + y2 + 2(58)x + 2(24) - 285 = 0


x2 + y2 + 116x + 48y - 285 = 0


The equation of the circle is x2 + y2 + 116x + 48y - 285 = 0.


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