if orthocenter and centroid of triangle are (-3,5) and (3,3) respectively, then find circumcentre

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Solution

$c . D i s \tan c e b e t w e e n c e n t r o i d a n d o r t h o c e n t e r i s e q u a l t o d o u b l e t h a t o f b e t w e e n c e n t r o i d a n d c i r c u m c e n t e r . A B = 2 B C S q u a r i n g b o t h s i d e s A B^{2} = 4 B C^{2} (- 3 - 3)^{2} + (5 - 3)^{2} = 40 = 4 {(3 - x)^{2} + (3 - y)^{2}} 10 = 9 + x^{2} - 6 x + 9 + y^{2} + 9 - 6 y x^{2} + y^{2} - 6 x - 6 y + 8 = 0 F r o m (1) p u t x = 12 - 3 y w e g e t, (12 - 3 y)^{2} + y^{2} - 6 (12 - 3 y) - 6 y + 8 = 0 y^{2} + 6 y + 8 = 0 y^{2} + 2 y + 4 y + 8 = (y + 4) (y + 2) = y = - 2 a n d y = - 4 x = 12 - 3 y s o, x = 12 + 6 = 18 a n d x = 12 + 12 = 24 s o c o o r d i n a t e s o f c i r c u m c e n t e r i s (18, - 2) a n d (24, - 4)$

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Write the coordinates of the circumcentre of a triangle whose centroid and orthoentre are at (3,3) and (-3,5) respectively.

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