Question

Let $C$ be the centre of the circle $x^2+y^2-x+2 y=\frac{11}{4}$ and $P$ be a point on the circle. A line passes through the point $C$, makes an angle of $\frac{\pi}{4}$ with the line CP and intersects the circle at the points $Q$ and $R$. Then the area of the triangle $PQR$ (in unit ${ }^2$ ) is :

• A. 2
• B. $2 \sqrt{2}$
• C. $8 \sin \left(\frac{\pi}{8}\right)$
• D. $8 \cos \left(\frac{\pi}{8}\right)$

Answer 1

Answer: (B)

$x^2+y^2-x+2 y=\frac{11}{4}$
$\left(x-\frac{1}{2}\right)^2+(y+1)^2=(2)^2$
Or $\triangle PQR$
$PR = QK \sin 2 \geq \frac{1}{3}$

$PQ = QR \cos 22 \frac{1}{2}$
$=4 \cos \frac{\pi}{8}$
$\text { As } \triangle PQR =\frac{1}{2} PR \times PQ$
$=\frac{1}{2}\left(4^2 \sin \frac{\pi}{6}\right)\left(4 \cos \frac{\pi}{8}\right)$
$=4 \sin \frac{\pi}{4}=\frac{4}{\sqrt{2}}=2 \sqrt{2}$