Question

Let $C$ be the centre of the circle $x^2+y^2-x+2y=\frac{11}{4}$ 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)

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

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