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Grade: 12th pass


how to prove that circumcenter of a triangle ABC with coordinates (x 1 , y 1 ), (x 2 y 2 ), (x 3 , y 3 ) is given by (x 1 sin2A + x 2 sin2B + x 3 sin2C)/ (sin2A+sin2B+sin2C), (y 1 sin2A + y 2 sin2B + y 3 sin2C)/ (sin2A+sin2B+sin2C)

how to prove that circumcenter of a triangle ABC with coordinates (x1, y1), (x2 y2), (x3, y3) is given by               (x1sin2A + x2sin2B + x3sin2C)/ (sin2A+sin2B+sin2C),    (y1sin2A + y2sin2B + y3sin2C)/ (sin2A+sin2B+sin2C)

11 months ago

Answers : (2)

25768 Points
Let co-ordinate of vertices of triangleABC (x1, y1) , (x2, y2) and (x3 , y3) respectively.
circumcentre :- point of intersection of perpendicular bisector of sides of traingle is known as circumcentre .
circumcentre (C) =
{(x1sin2A+x2sin2B+x3sin2C)/(sin2A+sin2B+sin2C), (y1sin2A+y2sin2B+y3sin2/(sin2A+sin2B+sin2C)}
11 months ago
Aditya Gupta
2075 Points
once again, aruns ans is wrong as he has not provided the proof that u were asking for. The actual proof is quite long and involves lengthy calculations, hence i am providing you the steps of the proof, without calculations:
  1. Let the mid pts of BC, CA, AB be D, E, F. obviously perpendiculars to BC, CA, AB at D, E, F intersect at the circumcentre O (by definition). we ll call the circumradius R.
  2. coordinates of D are ((x2+x3)/2, (y2+y3)/2)
  3. extend DO to intersect AB at P.
  4. by properties of circumcircle, DO= RcosA and OP= RcosC/cosB and PB= a/2cosB and PA= c – a/2cosB
  5. so, PO:OD= cosC:cosAcosB and BP:AP= a/2cosB: c – a/2cosB, or by sine rule, BP:AP= 1: 2cosBsinC/sinA – 1.
  6. by section formula, find the coordinates of P, and thence the reqd coordinates of O by again using section formula.
after all of these calculations, you will get the reqd coordinates. it personally took me half an hour to complete all the calculations. also, make sure not to make any mistakes during the entire prob. however, i would recommend u to simply memorise this formula, since the proof is well beyond the scope of jee. cheers!!
11 months ago
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