The point of intersection of the tangent x+√3y=8 with the circle x2+y2=16 is
A
(2√2,2√2)
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B
(2,2√3)
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C
(2√3,2)
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D
(2,−2√3)
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Solution
The correct option is B(2,2√3) Point form of tangent to circle xx1+yy1=16⋯(1) Tangent to the circle is x+√3y=8 Multiplying above equation by 2, we get 2x+2√3y=16⋯(2) Comparing equation (1) and (2), we get x1=2,y1=2√3