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QJEEM22Ma295e: JEE Main 2022 Mathematics - Class 11 JEEM22Ma295e Solution: Let be the centre of the circle and be a point on the circle. A …

Let CC be the centre of the circle x2+y2−x+2y=114x^{2}+y^{2}-x+2 y=\frac{11}{4} and PP be a point on the circle. A line passes through the point C\mathrm{C} , makes an angle of π4\frac{\pi}{4} with the line CP\mathrm{CP} and intersects the circle at the points QQ and RR . Then the area of the triangle PQRP Q R (in unit 2^{2} ) is :

Multiple Choice Options

  • A 2
  • B 2 2\sqrt2
  • C 8sin⁡(π8)8 \sin \left(\frac{\pi}{8}\right)
  • D 8cos⁡(π8)8 \cos \left(\frac{\pi}{8}\right)

Step-by-Step Verified Solution

JEE Main 2022 (Online) 28th July Morning Shift Mathematics - Circle Question 49 English Explanation

QR=2r=4QR = 2r = 4

P=(12+2cos⁡π4,−1+2sin⁡π4)P = \left( {{1 \over 2} + 2\cos {\pi \over 4}, - 1 + 2\sin {\pi \over 4}} \right)

=(12+2,−1+2)= \left( {{1 \over 2} + \sqrt 2 , - 1 + \sqrt 2 } \right)

Area of ΔPQR=12×4×2\Delta PQR = {1 \over 2} \times 4 \times \sqrt 2

=22= 2\sqrt 2 sq. units

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