Find the equation of the circle passing through the points :

(5, 7), (8, 1) and (1, 3)


Given that we need to find the equation of the circle passing through the points (5,7), (8,1) and (1,3).



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


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


Substituting (5,7) in (1), we get


52 + 72 + 2a(5) + 2b(7) + c = 0


25 + 49 + 10a + 14b + c = 0


10a + 14b + c + 74 = 0 ..... (2)


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


82 + 12 + 2a(8) + 2b(1) + c = 0


64 + 1 + 16a + 2b + c = 0


16a + 2b + c + 65 = 0 ..... (3)


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


12 + 32 + 2a(1) + 2b(3) + c = 0


1 + 9 + 2a + 6b + c = 0


2a + 6b + c + 10 = 0 ..... (4)


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


.


Substituting these values in (1), we get




3x2 + 3y2 - 29x - 19y + 56 = 0


The equation of the circle is 3x2 + 3y2 - 29x - 19y + 56 = 0.


2
1