Answer :

Given that we need to find the equation of the circle which passes through (3,7), (5,5) and has its centre on the line x - 4y = 1. ..... (1)



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


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


Substituting centre (- a, - b) in (1) we get,


(- a) - 4(- b) = 1


- a + 4b = 1


a - 4b + 1 = 0 ......(3)


Substituting (3,7) in (2), we get


32 + 72 + 2a(3) + 2b(7) + c = 0


9 + 49 + 6a + 14b + c = 0


6a + 14b + c + 58 = 0 ..... (4)


Substituting (5,5) in (2), we get


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


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


10a + 10b + c + 50 = 0 ..... (5)


Solving (3), (4) and (5) we get,


a = 3,b = 1,c = - 90


Substituting these values in (2), we get


x2 + y2 + 2(3)x + 2(1)y - 90 = 0


x2 + y2 + 6x + 2y - 90 = 0


The equation of the circle is x2 + y2 + 6x + 2y - 90 = 0.


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