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.
 A: Just a hint, are you sure integrals, for example, get out of hand?
Please look at the sketch below: your area is made up by four regions.
The quarter of a circle of radius $d$ marked with a blue cross, a rectangle (green cross), and two regions (marked by red crosses) under a portion of circular arc, easy and handy to parametrise

A: Let $p=k-a$ and $q=k-b$ and draw lines from $(a,b)$ to the three corners of the shaded region. Then the region breaks into two triangles and a sector.
The region's upper edge has length $\sqrt{d^2-q^2}+p=r$; the right edge has length $\sqrt{d^2-p^2}+q=s$. The area of the two triangles is then $\frac12(ps+qr)$. The angle subtended at $(a,b)$ by the right and upper edges is $\frac\pi2+\cos^{-1}\frac qd+\cos^{-1}\frac pd$, so the sector's angle is $\frac{3\pi}2-\cos^{-1}\frac qd-\cos^{-1}\frac pd=\theta$.
The sector's area is $\frac{\theta d^2}2=(3\pi/2-\cos^{-1}q/d-\cos^{-1}p/d)d^2/2$. Putting everything together, the total area is
$$\frac{p(\sqrt{d^2-p^2}+q)+q(\sqrt{d^2-q^2}+p)+(3\pi/2-\cos^{-1}q/d-\cos^{-1}p/d)d^2}2$$
A: You have 6 areas

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