Let C be the centre of the circle x^{2}+y^{2} | vedclass

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(A) The equation of the circle is $x^{2}+y^{2}-x+2y=\frac{11}{4}$.Completing the square,we get $(x-\frac{1}{2})^{2}+(y+1)^{2}=\frac{11}{4}+\frac{1}{4}

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$2$

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(A) The equation of the circle is $x^{2}+y^{2}-x+2y=\frac{11}{4}$.Completing the square,we get $(x-\frac{1}{2})^{2}+(y+1)^{2}=\frac{11}{4}+\frac{1}{4}+1 = 4 = 2^{2}$.Thus,the radius of the circle is $r=2$ and the centre is $C(\frac{1}{2}, -1)$.In $	riangle PQR$,$CP=CQ=CR=r=2$.The line through $C$ makes an angle of $\frac{\pi}{4}$ with $CP$. Let this line be $L$. $L$ intersects the circle at $Q$ and $R$. Thus,$\angle PCQ = \angle PCR = \frac{\pi}{4}$.In $	riangle PCQ$,$CP=CQ=2$ and $\angle PCQ = \frac{\pi}{4}$. The triangle is isosceles,so $\angle CPQ = \angle CQP = \frac{1}{2}(\pi - \frac{\pi}{4}) = \frac{3\pi}{8}$.The length of the chord $PQ = 2r \sin(\frac{\angle PCQ}{2}) = 2(2) \sin(\frac{\pi}{8}) = 4 \sin(\frac{\pi}{8})$.Similarly,$PR = 4 \sin(\frac{\pi}{8})$.The area of $	riangle PQR = \frac{1}{2} 	imes PQ 	imes PR 	imes \sin(\angle QPR)$.Since $\angle QPR = \angle QPC + \angle CPR = \frac{3\pi}{8} + \frac{3\pi}{8} = \frac{6\pi}{8} = \frac{3\pi}{4}$.Area $= \frac{1}{2} 	imes (4 \sin \frac{\pi}{8}) 	imes (4 \sin \frac{\pi}{8}) 	imes \sin(\frac{3\pi}{4}) = 8 \sin^{2}(\frac{\pi}{8}) 	imes \frac{1}{\sqrt{2}}$.Using $2 \sin^{2}(	heta) = 1 - \cos(2	heta)$,we have $8 \sin^{2}(\frac{\pi}{8}) = 4(1 - \cos \frac{\pi}{4}) = 4(1 - \frac{1}{\sqrt{2}}) = 4 - 2\sqrt{2}$.Area $= \frac{4 - 2\sqrt{2}}{\sqrt{2}} = 2\sqrt{2} - 2$. Wait,re-evaluating the area using base $QR$ and height $h$ from $P$ to $QR$:$QR = 2r \sin(\frac{\angle QCR}{2}) = 2(2) \sin(\frac{\pi}{4}) = 4 	imes \frac{1}{\sqrt{2}} = 2\sqrt{2}$.Height $h = r \cos(\frac{\pi}{4}) = 2 	imes \frac{1}{\sqrt{2}} = \sqrt{2}$.Area $= \frac{1}{2} 	imes QR 	imes h = \frac{1}{2} 	imes (2\sqrt{2}) 	imes \sqrt{2} = 2$.

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