Question

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 circles described on its sides as diameter is

• A. $\left( 3, - \frac{1}{\sqrt{3}} \right)$
• B. $\left( 3, - \sqrt{3} \right)$
• C. $\left( 2, - \frac{1}{\sqrt{3}} \right)$
• D. $(2, - \sqrt{3})$

Answer 1

Answer: (C)

$\left( 2, - \frac{1}{\sqrt{3}} \right)$

Answer 2

Radical centre of the circles described on the sides of a triangle as diameters is the orthocenter of the triangle

\ OD = 2

DH = – BD tan $\frac{\pi}{6} = –\frac{1}{\sqrt{3}}$

\ coordinates of H are $\left( 2, - \frac{1}{\sqrt{3}} \right)$