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  4. Let C be the centre of the circle x2+y2-x+2 y=(11/4) and P be a point on the circle. A line passes through the point C, makes an angle of (π/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 :

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Q. 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 :

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Solution:

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