The coodiantes of two points on the circle $x^{2} + y^{2} - 12 x - 16 y + 75 = 0,$ one nearest to the origin and the other farthest from it, are

(3,4),(9,12)

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(3,2),(9,12)

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(-3,4),(9,12)

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(3,4),(9,-12)

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Solution

The correct option is A (3,4),(9,12)
Let O be the origin and A,B be the nearest and farthest point
Centre c=(6,8) $r a d i u s = 5 = A C o c = \sqrt{36 + 64} = 10 O A = 5 ∵ O A : A C = 5 : 5 = 1 : 1 A i s m i d p o i n t 0 f O C i . e, (3, 4) c o o r d i a n t e B b e (h, k)^{'} c^{'} i s t h e m i d p o i n t 0 f A B ∴ h = 9, k = 12 ∴ B (9, 12)$

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