Question

show that the points (9, 1),(7,9),(-2,12)and(6,10) are concyclic

Answer 1

The base equation for a circle centered at (h,k) with radius r is:

   (x - h)2 + (y - k)2 = r2

If the points are concyclic, they all lie on the graph for the same circle.

Thus, all points are solutions to a single equation of the circle 

WE can plug each point into the base equation, and obtain a system of equationd

FOR POINT (-2,10)

     Equation 1:  (-2 - h)2 + (10 - k)2 = r2

FOR POINT (1,11)

     Equation 2:  (1 - h)2 + (11 - k)2 = r2

FOR POINT (6,10)

     Equation 3:  (6 - h)2 + (10 - k)2 = r2

FOR POINT (9,7)

     Equation 4:  (9 - h)2 + (7 - k)2 = r2


From Equation 1 and 2 above, since they have 2 different expressions equivalent to r2, we can construct an equation involving variables x and h.

     r2 = r2   

     (-2 - h)2 + (10 - k)2 = (1 - h)2 + (11 - k)2

     (-2 - h)(-2 - h) + (10 - k)(10 - k) = (1 - h)(1 - h) + (11 - k)(11 - k)

     (4 + 2h + 2h + h2) + (100 -10k - 10k + k2) = (1 - h - h + h2) + (121 - 11k - 11k + k2) 

      h2 + 4h + 4 + k2 - 20k + 100 = h2-2h + 1 + k2 -22k + 121 

    We can subtract h and k from both sides.

      4h + 4 - 20k + 100 = -2h + 1 -22k + 122

      4h -20k + 104 = -2h - 22k + 122

    Move all h and k variables to the left ... move all constants to the right

      6H + 2K = 18

SYSTEM EQUATION 1:   6h + 2k = 18      

From equation 3 and 4 above, we go through the same process to obtain a second system equation.

       r2 = r2

       (6 - h)2 + (10 - k)2 = (9 - h)2 + (7 - k)2

       (6 - h)(6 - h) + (10 - k)(10 - k) = (9 - h)(9 - h) + (7 - k)(7 - k)

       36 - 6h - 6h + h2 + 100 - 10k - 10k + k2 = 81 - 9h - 9h + h2 + 49 = 7k - 7k + k2

       h2 - 12h + 36 + k2 - 20k + 100 = h2 - 18h + 81 + k2 - 14k + 49 

    Simplify, and subtract h2 and k2from both sides

       -12h - 20k + 136 = -18h - 14k + 130

      Move the h and k variables to the left, and the constants to the right

        -12h + 18h - 20k + 14k = 130 - 136

        6h - 6k = -6

SYSTEM EQUATION 1:  6h - 6k = -6

      6h +2k = 18

      6h - 6k = -6

Multiply the second equation by -1, we have ... -6h + 6k = 6

Adding the 2 equations, we have

 6h + 2k = 18

  -6h + 6k =  6

          8k =24

             k = 3

Using substitution of k=3 into equation 6h + 2k = 18,
   
6h + 2(3) = 18

    6h + 6 = 18

    6h = 12

    h = 2

Now, use one of the points on the circle ... (1, 11) and (h,k) ... (2,3), to find the radius

   (1 - 2)2 + (11 - 3)2 = r2

   1 + 64 = r2

   65 = r2

   r = √65
Thus the equation of our circle is:
   (x - 2)2 + (y - 3)2 = (√65)2