show that (2,1) is the centre of the circumcircle of the triangle whose vertices are (-3,-9),(13,-1) and (-9,3).
Answers
Given : (2,1) is the centre of the circumcircle of the triangle whose vertices are (-3,-9),(13,-1) and (-9,3).
To find : verify
Solution :
centre of the circumcircle of the triangle is at equal distance from all the vertex of triangle
centre of the circumcircle ( 2 ,1 )
Vertex are (-3,-9),(13,-1) and (-9,3).
Distance of ( 2 ,1 ) from (-3,-9)
= √(2 - (-3))² + (1 - (-9))²
= √5² + 10²
= √25 + 100
= √125
= 5√5
Distance of ( 2 ,1 ) from (13,-1)
= √(2 - (13))² + (1 - (-1))²
= √(-11)² + (2)²
= √121 + 4
= √125
= 5√5
Distance of ( 2 ,1 ) from (-9,3)
= √(2 - (-9))² + (1 - (3))²
= √11² + (-2)²
= √121 + 4
= √125
= 5√5
(2,1) is at equal distance 5√5 from all the vertex hence
(2,1) is the centre of the circumcircle of the triangle whose vertices are (-3,-9),(13,-1) and (-9,3).
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