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.

Question

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 :

Accepted Answer

$QR = 2r = 4$

$$P = \left( {{1 \over 2} + 2\cos {\pi  \over 4}, - 1 + 2\sin {\pi  \over 4}} \right)$$

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

Area of $$\Delta PQR = {1 \over 2} \times 4 \times \sqrt 2 $$

$$ = 2\sqrt 2 $$ sq. units

Solutions · 1