Question

show that pts. (9,1) (7,9) ,(-2,12) ,(6,10) are concyclic

Answer 1

 

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 k2 from both sides