On the line segment joining (1, 0) and (3, 0) an equilateral triangle is drawn having its…
Mathematics · JEE Advanced · NTA Exams — Co-ordinate Geometry
On the line segment joining (1, 0) and (3, 0) an equilateral triangle is drawn having its vertex in the fourth quadrant, then radical centre of the circle described on its sides as diameter is
- \(\left(3,-\frac{1}{\sqrt{3}}\right)\)
- \((3,-\sqrt{3})\)
- \(\left(2,-\frac{1}{\sqrt{3}}\right)\)
- \((2,-\sqrt{3})\)
Answer
(C) (2,- 1 3 )
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