Let C be the centre of the circle x 2 + y 2 - x + 2 y = 11 4 and P be a point on the circle. A line passes…

Let C be the centre of the circle x2+y2-x+2y=114 and P be a point on the circle. A line passes through the point C, makes an angle of π4 with the line CP and intersects the circle at the points Q and R. Then the area of the triangle PQR (in unit2) is
  1. 2
  2. 22
  3. 8sinπ8
  4. 8cosπ8

Solution

Given,

x2+y2-x+2y=114

On rearranging terms we get,

x-122+y+12=22

Now given in PQRPCR=π4 so PQR=2212 by circle property, radius is 2 so RC=QC=2

Now, PR=QRsin2212=4sinπ8

And PQ=QRcos2212=4cos2212

Now area of ΔPQR=12PR×PQ

=124sinπ84cosπ8

=4×2sinπ8cosπ8

=4sinπ4=42=22

Asked in: JEE Main 2022 (28 Jul Shift 1)

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