If coordinates of the centre and one end of a diameter of a circle are (7,3) and (5,−7) respectively, then the coordinates of the other end of the diameter are
A
(6,−2)
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B
(9,13)
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C
(12,−4)
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D
(−2,6)
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Solution
The correct option is B(9,13) Let A≡(5,−7),C≡(7,3)
Let B(x,y) be the other end of the diameter.
We know that in a circle, the centre bisects the diameter, so C is midpoint of AB. (7,3)=(5+x2,−7+y2)⇒x=9,y=13∴B≡(9,13)