JEE MAIN - Mathematics (2022 - 28th July Morning Shift - No. 10)
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 $$\mathrm{C}$$, makes an angle of $$\frac{\pi}{4}$$ with the line $$\mathrm{CP}$$ and intersects the circle at the points $$Q$$ and $$R$$. Then the area of the triangle $$P Q R$$ (in unit $$^{2}$$ ) is :
2
2$$\sqrt2$$
$$8 \sin \left(\frac{\pi}{8}\right)$$
$$8 \cos \left(\frac{\pi}{8}\right)$$
Explanation
_28th_July_Morning_Shift_en_10_1.png)
$$QR = 2r = 4$$
$$P = \left( {{1 \over 2} + 2\cos {\pi \over 4}, - 1 + 2\sin {\pi \over 4}} \right)$$
$$ = \left( {{1 \over 2} + \sqrt 2 , - 1 + \sqrt 2 } \right)$$
Area of $$\Delta PQR = {1 \over 2} \times 4 \times \sqrt 2 $$
$$ = 2\sqrt 2 $$ sq. units
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