Show that the points ( 9 , 1 ) , ( 7 , 9 ) , ( − 2 , 12 ) , ( 6 , 10 ) are concyclic.

Question

Show that the points   (  9  ,  1  )  ,  (  7  ,  9  )  ,  (  −  2  ,  12  )  ,  (  6  ,  10  ) are concyclic.

livin4him777lf · Accepted Answer

Step 1Here&#039;s a completely different approach. We regard your 4 points as complex numbers      z    1    =  9  +  i  ,            z    2    =  7  +  9  i  ,            z    3    =  −  2  +  12  i  ,            z    4    =  6  +  10  iand now we calculate the cross ratio of these points. It is a basic fact about the cross ratio that the cross ratio is real if and only if the points are either colinear or concyclic. They are clearly not colinear, just by looking at them, so if the cross ratio  C  (      z    1    ,      z    2    ,      z    3    ,      z    4    )  =            (              z        3            −              z        1            )      (              z        4            −              z        2            )              (              z        3            −              z        2            )      (              z        4            −              z        1            )      is real, then we&#039;re done. Plugging everything in, we have:  C  (      z    1    ,      z    2    ,      z    3    ,      z    4    )  =            (      (      −      2      +      12      i      )      −      (      9      +      i      )      )      (      (      6      +      10      i      )      −      (      7      +      9      i      )      )              (      (      −      2      +      12      i      )      −      (      7      +      9      i      )      )      (      (      6      +      10      i      )      −      (      9      +      i      )      )        =            (      −      11      +      11      i      )      (      −      1      +      i      )              (      −      9      +      3      i      )      (      −      3      +      9      i      )      which we can simplify down to  =      11    3              −      2      i              −      10      i      which is evidently real, since the i&#039;s cancel.

freakygirl838w · Accepted Answer

Step 1You can use Ptolemy&#039;s inequality to check if the points are concyclic.Define   A  =  (  9  ,  1  )  ,  B  =  (  7  ,  9  )  ,  C  =  (  −  2  ,  12  )  , and   D  =  (  6  ,  10  ) . The inequality states that the points A, B, C, and D are concyclic if  A  B  ⋅  C  D  +  A  D  ⋅  B  C  =  A  C  +  B  DBy solving for the distances, we have  A  B  =  2      17    ,  C  D  =  2      17    ,  A  C  =  11      2    ,  B  D  =      2    ,  A  D  =  3      10   and   B  C  =  3      10   . You can check that the inequality becomes an equality, which means they are concyclic.

Show that the points ( 9 , 1 ) , ( 7 , 9 ) , ( −<!

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