Question

A circle $C$ of radius $1 $ is inscribed in an equilateral triangle $PQR$. The points of contact of $C$ with the sides $P Q, Q R, R P$ are $D, E, F$, respectively. The line $P Q$ 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 $C E$ is $\frac{x-\sqrt{3}}{-\frac{\sqrt{3}}{2}}=\frac{y-1}{\frac{1}{2}}=1$
$\Rightarrow E \equiv\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right)$