## show that (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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## 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 :verifySolution :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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