JEE Advanced · Mathematics · 9. Straight Lines
Paragraph:
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
- 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 & Solution
Correct Answer
(A)
\(\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right),(\sqrt{3}, 0)\)
Step-by-step Solution
Detailed explanation
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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