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Question:
find the equation of the circle whose center lie on the line x-4y=1 and which passes through the point (3,7) and (5,5).
Answer:

The general form of the circle is

x2 +y2 + 2gx + 2fy + c = 0 ...............1

Centre of the circle is (-g , -f)

Now center lies on the line x - 4y = 1

So  -g - 4*(-f) = 1

=> -g + 4f = 1 ..............2

Again the circle passes through the points (3,7) and (5,5) 

      32 + 72 + 2*g*3 + 2*f*7 + c = 0

=> 9 + 49 + 6g + 14f  + c = 0

=> 58 + 6g + 14f  + c = 0

=> 6g + 14f  + c = -58 .........3

and

      52 + 52 + 2*g*5 + 2*f*5 + c = 0

=> 25 + 25 + 10g + 10f  + c = 0

=> 50 + 10g + 10f  + c = 0

=> 10g + 10f  + c = -50 ..........4

Now equation 3 - equation 4, we get

-4g + 4f = -8

=> -g + f = -2 .......5

Now equation 5 - equation 2, we get

-3f = -3

=>f = 1

From equation 5

-g + 1 = -2

=> -g = -2 - 1

=> -g = -3

=> g = 3

Put value of g and f in equation 3, we get

6*3 + 14*1 + c = -58

=> 18 + 14 + c = -58

=> 32 + c = -58

=> c = -58 - 32

=> c  = -90

from equation 1, we get

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

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

This is the required equation of the circle. 

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