Dear Student, Centre is the midpoint of AB Let the centre be CC = 3+ 92,1+102 = 6,112Let the diameter of this new circle be ED where it cuts AB Given, ED = 13AB12ED = 12(13AB)EC = 13ACECAC = 13ECAC - EC = 13-1ECAE = 12AE : EC = 2:1Let E = (a,b)a = 2 6 + 1 32+1 = 5b =2 112 + 1 12+1 = 4E = (5,4)Similarly BD:DC = 2:1Let D = (x,y)x = 2 6 + 1 92+1 = 7y =2 112 + 1 102+1 = 7D = (7,7) D and E are the required point Regards,

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Dear Student,

C e n t r e i s t h e m i d p o i n t o f A B . L e t t h e c e n t r e b e C C = (\frac{3 + 9}{2}, \frac{1 + 10}{2}) = (6, \frac{11}{2}) L e t t h e d i a m e t e r o f t h i s n e w c i r c l e b e E D w h e r e i t c u t s A B . G i v e n, E D = \frac{1}{3} A B \frac{1}{2} E D = \frac{1}{2} (\frac{1}{3} A B) E C = \frac{1}{3} A C \frac{E C}{A C} = \frac{1}{3} \frac{E C}{A C - E C} = \frac{1}{3 - 1} \frac{E C}{A E} = \frac{1}{2} A E : E C = 2 : 1 L e t E = (a, b) a = \frac{2.6 + 1.3}{2 + 1} = 5 b = \frac{2 . \frac{11}{2} + 1.1}{2 + 1} = 4 E = (5, 4) S i m i l a r l y B D : D C = 2 : 1 L e t D = (x, y) x = \frac{2.6 + 1.9}{2 + 1} = 7 y = \frac{2 . \frac{11}{2} + 1.10}{2 + 1} = 7 D = (7, 7)

D and E are the required point

Regards,

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