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  • Let  be the centre of the circle <img alt="\mathrm{ x^{2}+y^{2}-x+2 y=\frac{11}{4}}" src="/latex-image/?%5Cmathrm%7B%20x%5E%7B

Let \mathrm{C} be the centre of the circle \mathrm{ x^{2}+y^{2}-x+2 y=\frac{11}{4}} and \mathrm{ \mathrm{P}} be a point on the circle. A line passes through the point \mathrm{C}, makes an angle of \mathrm{ \frac{\pi}{4}} with the line \mathrm{ \mathrm{CP}} and intersects the circle at the points \mathrm{ Q} and \mathrm{ R}. Then the area of the triangle \mathrm{ P Q R} (in unit2 ) is :

Option: 1

2


Option: 2

2 \sqrt{2}


Option: 3

8 \sin \left(\frac{\pi}{8}\right)


Option: 4

8 \cos \left(\frac{\pi}{8}\right)


Answers (1)

best_answer

\begin{aligned} & \mathrm{x^{2}+y^{2}-x+2 y=\frac{11}{4}} \\ & \mathrm{\left(x-\frac{1}{2}\right)^{2}+(y+1)^{2}=2^{2}} \\ &\text { or } \mathrm{ \triangle PQ R }\\ & \mathrm{P R=2 k \times \sin 2 \geq \frac{1}{3}} \\ & \mathrm{=4.6 \sin \frac{\pi}{8}} \\ & \mathrm{P Q=Q R \cos 22 \frac{1}{2}} \\ & \mathrm{=4 \cos \frac{\pi}{8}} \\ &\text { As } \mathrm{\triangle P 2 R=\frac{1}{2} P R \times P Q} \\ & \mathrm{=\frac{1}{2}\left(4^{2} \sin \frac{\pi}{6}\right)\left(4 \cos \frac{\pi}{8}\right)} \\ &\mathrm{4 \sin \frac{\pi}{4}=\frac{4}{\sqrt{2}}=2\sqrt{2}}\end{aligned}

 

 

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Ritika Harsh

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