Find the area of a circle part A circle is centred at $(a,b)$ and has radius $d$. The value $k$ is such that the lines $y =k $ and $x=k$ both intersect the circle twice. Moreover, $a,b<k$, so that the point $(k,k)$ is inside the circle, but also above, and to the right of, the point $(a,b)$. This setup is illustrated below.

Given this information, what is the area of the shaded region?
I tried to do this geometrically, but couldn't work out the appropriate angles. Then I attempted it by using integrals, but this also became very complicated very quickly. It seems like there should be a neat way of seeing what this area is, but I can't seem to find it.

You have 6 areas

$P_1=\frac{d^2\pi}{4}$
$P_2=\arcsin(\frac{k-a}d)\frac{d^2}{2}$
$P_3=\arcsin(\frac{k-b}d)\frac{d^2}{2}$
$P_4=\cos\arcsin(\frac{k-a}d)\frac{d(k-a)}{2}$
$P_5=\cos\arcsin(\frac{k-b}d)\frac{d(k-b)}{2}$
$P_6=(k-a)(k-b)$
