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A circle $C$ of radius 1 is inscribed in an equilateral $\triangle P Q R$. 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 $P Q$.Question:
Points $E$ and $F$ are given by
Options:
A circle $C$ of radius 1 is inscribed in an equilateral $\triangle P Q R$. 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 $P Q$.Question:
Points $E$ and $F$ are given by
Solution:
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Verified Answer
The correct answer is:
$\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right),(\sqrt{3}, 0)$
$\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right),(\sqrt{3}, 0)$
Slope of line joining centre of circle to point $D$
$$
=\frac{\frac{3}{2}-1}{\frac{3 \sqrt{3}}{2}-\sqrt{3}}=\frac{1}{\sqrt{3}}
$$
[makes an angle $30^{\circ}$ with $X$-axis]
$\therefore$ Points $E$ and $F$ will make angle $150^{\circ}$ and $-90^{\circ}$ with $X$-axis.
$\therefore E$ and $F$ are given by and
$\therefore$
$$
\begin{aligned}
\frac{x-\sqrt{3}}{\cos 150^{\circ}} & =\frac{y-1}{\sin 150^{\circ}}=1 \\
\frac{x-\sqrt{3}}{\cos \left(-90^{\circ}\right)} & =\frac{y-1}{\sin \left(-90^{\circ}\right)}=1 \\
E & =\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right) \\
and
F & =(\sqrt{3}, 0)
\end{aligned}
$$
$$
=\frac{\frac{3}{2}-1}{\frac{3 \sqrt{3}}{2}-\sqrt{3}}=\frac{1}{\sqrt{3}}
$$
[makes an angle $30^{\circ}$ with $X$-axis]
$\therefore$ Points $E$ and $F$ will make angle $150^{\circ}$ and $-90^{\circ}$ with $X$-axis.
$\therefore E$ and $F$ are given by and
$\therefore$
$$
\begin{aligned}
\frac{x-\sqrt{3}}{\cos 150^{\circ}} & =\frac{y-1}{\sin 150^{\circ}}=1 \\
\frac{x-\sqrt{3}}{\cos \left(-90^{\circ}\right)} & =\frac{y-1}{\sin \left(-90^{\circ}\right)}=1 \\
E & =\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right) \\
and
F & =(\sqrt{3}, 0)
\end{aligned}
$$
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