1. Tardigrade
  2. Question
  3. Mathematics
  4. Consider 3 non collinear point A (9,3) ; B (7,-1) and C (1,-1). Let P ( a , b ) be the centre and ' R ' is the radius of the circle 'S' passing through A, B, C. Also H ( barx, bary) are the coordinates of the orthocentre of the triangle ABC whose area be denoted by triangle. The value of (a+b+R) equals

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Q. Consider 3 non collinear point $A (9,3) ; B (7,-1)$ and $C (1,-1)$. Let $P ( a , b )$ be the centre and ' $R$ ' is the radius of the circle 'S' passing through A, B, C. Also H ( $\bar{x}, \bar{y})$ are the coordinates of the orthocentre of the triangle $ABC$ whose area be denoted by $\triangle$.
The value of $(a+b+R)$ equals

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Solution:

Since $P(a, b)$ is the circumcentre
$\therefore (a-7)^2+(b+1)^2=(a-9)^2+(b-3)^2 $ ....(1)
$(a-7)^2+(b+1)^2=(a-1)^2+(b+1)^2$ .....(2)
$\text { solving }(1) \text { and }(2)$
$ a=4 \text { and } b=3 $
$\therefore R=\sqrt{(4-7)^2+(3+1)^2}=5$
$\therefore a+b+c=4+3+5=12 $