1. Find the centre and radius of the circle

Given: $x^2+y^2-x+2y=\frac{11}{4}$

Complete squares: $x^2-x+y^2+2y=\frac{11}{4}$ $\left(x-\frac12\right)^2-\frac14+(y+1)^2-1=\frac{11}{4}$ $\left(x-\frac12\right)^2+(y+1)^2=\frac{11}{4}+\frac14+1=4$

So the circle has:

• Centre $C=\left(\frac12,-1\right)$
• Radius $r=2$

2. Understand the geometry

Point $P$ is on the circle, so: $CP=r=2$

A line through $C$ makes an angle of $\frac{\pi}{4}$ with line $CP$, and intersects the circle at $Q$ and $R$.

Since the line passes through the centre, $QR$ is a diameter of the circle. Hence: $QR=2r=4$

Also, the angle between $CP$ and the line $CQ$ is $\frac{\pi}{4}$. Therefore the central angle between $CP$ and $CQ$ is $45^\circ$, and between $CP$ and $CR$ is $180^\circ-45^\circ=135^\circ$.

3. Compute the area of triangle $PQR$

Take $QR$ as the base. Since $QR$ is a line through the centre, the perpendicular distance from $P$ to line $QR$ is the component of radius $CP$ perpendicular to $QR$.

Because angle between $CP$ and $QR$ is $\frac{\pi}{4}$,

$\text{distance from }P\text{ to }QR = CP\sin\frac{\pi}{4}$ $=2\cdot \frac{\sqrt2}{2}=\sqrt2$

So area is: $\Delta_{PQR}=\frac12\times QR\times \text{height}$ $=\frac12\times 4\times \sqrt2$ $=2\sqrt2$

4. Check options

• A: $2$ ❌
• B: $2\sqrt2$ ✅
• C: $8\sin\left(\frac{\pi}{8}\right)$ ❌
• D: $8\cos\left(\frac{\pi}{8}\right)$ ❌

Thus the correct option is B.

5. Compare with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.