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Question:
Find the equation of the circle passing through the points (3,4) (3,2) (1,4).
Answer:

The general equation of circle is

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

The circle passes througt the points (3,4) (3,2) (1,4)

Take point (3, 4)

      32 + 42 + 2g*3 + 2f*4 + c = 0

=> 9 + 16 + 6g + 8f + c = 0

=> 25 + 6g + 8f + c = 0

=> 6g + 8f + c = -25 ..............2

Take point (3, 2)

      32 + 22 + 2g*3 + 2f*2 + c = 0

=> 9 + 4 + 6g + 4f + c = 0

=> 13 + 6g + 4f + c = 0

=> 6g + 4f + c = -13 ..............3

Take point (1, 4)

      12 + 42 + 2g*1 + 2f*4 + c = 0

=> 1 + 16 + 2g + 8f + c = 0

=> 17 + 2g + 8f + c = 0

=> 2g + 8f + c = -17 ..............4

Now equation 2 - equation 3, we get

     4f = -12

=> f = -12/4

=> f = -3

Again equation 2 - equation 4, we get

     4g = -8

=> g = -8/4

=> g = -2

Put value of f and g in equation 2, we get

      6*(-2) + 8*(-2) + c = -25

=> -12 - 16 + c = -25

=> -28 + c = -25

=> c = 28 - 25

=> c = 3

Put value of f, g and c in equation 1, we get

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

x2 + y2 - 4x - 6y + 3 = 0

This is the required equation of circle.

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