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. The line $PQ$ is given by the equation $\sqrt{3}x+y-6=0$ and the point $D$ is $\left(\frac{3\sqrt{3}}{2},\frac{3}{2}\right)$. Further, it is given that the origin and the centre of $C$ are on the same side of the line $PQ$.
Points $E$ and $F$ are given by

• A. $\left(\frac{\sqrt{3}}{2},\frac{3}{2}\right),(\sqrt{3},0)$
• B. $\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right),(\sqrt{3},0)$
• C. $\left(\frac{\sqrt{3}}{2},\frac{3}{2}\right),\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)$
• D. $\left(\frac{3}{2},\frac{\sqrt{3}}{2}\right),\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)$

Answer 1

Answer: (A)

Since the radius of the circle is $1$ and $C(\sqrt{3},1)$, coordinates of $F\equiv (\sqrt{3},0)$
Equation of $CE$ is $\frac{x-\sqrt{3}}{-\frac{\sqrt{3}}{2}}=\frac{y-1}{\frac{1}{2}}=1$
$\Rightarrow E\equiv \left(\frac{\sqrt{3}}{2},\frac{3}{2}\right)$