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JEE Mains · Maths · STD 11 - 10.1 circle and system of circle

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 \(C\), makes an angle of \(\frac{\pi}{4}\) with the line \(C P\) and intersects the circle at the points \(Q\) and \(R\). Then the area of the triangle \(P Q R\) (in unit \({ }^{2}\) ) is.

  1. A \(2\)
  2. B \(2 \sqrt{2}\)
  3. C \(8 \sin \left(\frac{\pi}{8}\right)\)
  4. D \(8 \cos \left(\frac{\pi}{8}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(2 \sqrt{2}\)

Step-by-step Solution

Detailed explanation

\(x ^{2}+ y ^{2}- x +2 y =\frac{11}{4}\) \(\left( x -\frac{1}{2}\right)^{2}+( y +1)^{2}=(2)^{2}\) Or \(\triangle PQR\) \(PR = QK \sin 2 \geq \frac{1}{3}\) \(=4 \cdot 6 \sin \frac{\pi}{8}\) \(PQ = QR \cos 22 \frac{1}{2}\) \(=4 \cos \frac{\pi}{8}\) As…