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2015-04-01T21:35:37+05:30

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The only point which will be equidistant from the three given points is the circum-center O of the triangle formed by the points A (5, 3),  B(5, -5) and C (1, -5).

It lies on the intersection of the perpendicular bisectors of the sides AB, BC and CA.

Mid point of AB = D = [ (5+5)/2, (3-5)/2 ] = (5, -1)
Mid point of BC = E = [ (5+1)/2 , (-5-5)/2 ] = (3, -5)

slope of AB = (-5 -3)/ (5-5) = ∞
     =>  Slope of perpendicular bisector DO of AB = 0
equation of DO :  y = 0 *x + c
       as D lies on it,    -1 = c    =>    DO is :  y = -1.

slope of BC =  (-5 +5) / (1 -5) = 0
     => slope of the perpendicular bisector EO of BC is = ∞.
   => equation of  EO is :    x = c
           as  E lies on EO,      3 = c.
       =>  EO is x = 3.

The intersection of DO and EO is the circumcenter O :  (3, -1).
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another way:
 
   let the point equidistant be O (x, y).
Hence,  OA² = OB² = OC²
   (x -5)² + (y-3)² = (x-5)² + (y+5)²  =  (x -1)² + (y + 5)²

From the first equality, we get:
         y - 3 = - (y + 5)             as  y - 3  ≠  y + 5
         2 y = -2  =>  y = -1

From the second equality we get : 
        x - 5 = - (x - 1)               as  x - 5 ≠ x - 1
          2 x = 6          =>  x = 3
 So  (3 , -1) is the equidistant point.

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