Let C be the centre of the circle x^2+y^2-x+2y=114 and P be a point on the circle.

Question

Let $C$ be the centre of the circle $x^2+y^2-x+2y=\dfrac{11}4$ and $P$ be a point on the circle. A line passes through the point $C$, makes an angle of $\dfrac\pi4$ with the line $CP$ and intersects the circle at the points $Q$ and $R$. Then the area of the triangle $PQR$ (in unit$^2$) is:

Accepted Answer

Centre $\left(\frac12,-1\right)$, $r^2=\frac14+1+\frac{11}4=4\Rightarrow r=2$. Since the line through $C$ meets the circle at $Q,R$, $QR$ is a diameter: $QR=4$.

Height of $P$ above line $QR$ is $r\sin\frac\pi4=2\cdot\frac{\sqrt2}2=\sqrt2$.

$$\text{Area}=\frac12\cdot4\cdot\sqrt2=2\sqrt2.$$

Solutions · 1