Question

A circle $C$ of radius $1$ is inscribed in an equilateral triangle $PQR$. The points of contact of $C$ with the sides $PQ,QR,RP$ are $D,E,F$, respectively. Then line $PQ$ is given by the equation $\sqrt{3}x+y-6=0$ and the point $D$ is $(\frac{3\sqrt{3}}{2},\frac{3}{2})$. Further, it is given that the origin and the centre of $C$ are on the same side of the line $PQ$.

Equations of the sides QR, RP are

• A. $y=\frac{2}{\sqrt{3}}x+1,y=-\frac{2}{\sqrt{3}}x-1$
• B. $y=\frac{x}{\sqrt{3}},y=0$
• C. $y=\frac{\sqrt{3}}{2}x+1,y=-\frac{\sqrt{3}}{2}x-1$
• D. $y=\sqrt{3}x,y=0$

Answer 1

The correct option is D $y=\sqrt{3}x,y=0$