## show that (2,1) is the centre of the circumcircle of the triangle whose vertices are (-3,-9),(13,-1) and (-9,3).

Question

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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18 hours 2021-10-13T09:48:55+00:00 2 Answers 0 views 0

1. 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).