A(θ) = Area of quadrilateral – area of sector

$=\cot \frac{\mathrm{\theta }}{2}-\frac{\mathrm{\pi }-\mathrm{\theta }}{2}$

$\text{ if }\mathrm{\theta }=\frac{\mathrm{\pi }}{4}, \mathrm{A}\left(\frac{\mathrm{\pi }}{4}\right)=\cot \frac{\mathrm{\pi }}{8}-\frac{3\mathrm{\pi }}{8}=\sqrt{2}+1-\frac{3\mathrm{\pi }}{8}$.

${\mathrm{m}}_{\mathrm{CQ}}=\frac{\mathrm{y}-1}{\mathrm{x}-1}=\mathrm{cot\theta }$

mPR = – tan θ

equation of PR is

$\mathrm{y}-(1+\mathrm{cos\theta })=-\mathrm{tan\theta }(\mathrm{x}-(1+\mathrm{sin\theta }\left)\right)$

$\mathrm{y}+\mathrm{xtan\theta }=(1+\mathrm{cos\theta })+\mathrm{tan\theta }(1+\mathrm{sin\theta })=\frac{\mathrm{cos\theta }(1+\mathrm{cos\theta })+\mathrm{sin\theta }(1+\mathrm{sin\theta })}{\mathrm{cos\theta }}$

$\mathrm{y}+\mathrm{xtan\theta }=\frac{1+\mathrm{cos\theta }+\mathrm{sin\theta }}{\mathrm{cos\theta }}$

x sin θ + y cos θ = (cos θ + sin θ + 1)

For the circle with C, using parametric

x – 1 = sin θ
y – 1 = cos θ

$\left.\Rightarrow \begin{array}{l}\mathrm{x}=1+\mathrm{sin\theta }\\ \mathrm{y}=1+\mathrm{cos\theta }\end{array}\right\}\Rightarrow (1+\mathrm{sin\theta },1+\mathrm{cos\theta })$.