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

Question

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 (A) 3 (B) 12 (C) 13 (D) None

Tardigrade Admin · Accepted Answer

Since P(a, b) is the circumcentre ∴ (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 ∴ R=√(4-7)2+(3+1)2=5 ∴ a+b+c=4+3+5=12

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 ( xˉ,yˉ​) are the coordinates of the orthocentre of the triangle ABC whose area be denoted by △. The value of (a+b+R) equals

3

12

13

None

Since P(a,b) is the circumcentre ∴(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) solving (1) and (2) a=4 and b=3 ∴R=(4−7)2+(3+1)2​=5 ∴a+b+c=4+3+5=12

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