Formula to Calculate Each Pie Angle Where the Intercepted Arc is NOT the Center Point of the Circle AND All Slices are Equal Sized 
I saw this picture titled "How to Start a Fight at Thanksgiving" and it made me laugh and then it made me wondered how to cut a pie into (N) number of pieces of equal surface area, but the central point of arc interception (C) is NOT the center point, instead it's located somewhere else inside the pie at coordinates X,Y.
Is there a formula to calculate the different angles so that each slice has the same surface area?
For discussion assume (r) Radius 4.5", (n) Number of slices is 6, (c) central point of arc interception is 1" moved to the left (west) of the true center point of the circle and 1.5" towards the top (north).
$\frac{\pi r^2}{n} = $ ~10.603 sq.inches for each slice, so what would be the different angles so that each slice equals ~10.603 sq. inches?
Assumption: the first single cut is the shortest line possible from the common point to the perimeter and were dealing with 3 or more (n) number of slices.
I thought this would be a fun Thanksgiving puzzle to solve. Thanks for playing.
 A: Let’s place the point at which all of the cuts converge at the origin and the center of the circular pie at $(h,0)$ so that the circle can be parameterized as $x=h+r\cos t$, $y=r\sin t$. The parameter $t$ represents the angle at the center of the pie. If $\Gamma$ is the arc of the circle that goes from $t_1$ to $t_2$ then the area of the slice is $$\frac12\int_\Gamma x\,dy-y\,dx = \frac r2\int_{t_1}^{t_2}r+h\cos t\,dt = \frac r2\left(r(t_2-t_1)+h(\sin t_1-\sin t_0)\right).$$ If we want $n$ equally-sized slices, this area must be equal to $\pi r^2/n$, which leads to the equation $$rt_2+h\sin t_2 = \frac{2\pi r}n+rt_1+h\sin t_1.$$ If we fix $t_1$, this can be solved for $t_2$. Unfortunately, there’s no closed-form solution, but you can get a numerical approximation good enough for making the slices.  
Taking your example, $h=\sqrt{1^2+1.5^2}\approx1.803$ and the area of each slice is approximately $10.603.$ The first cut is at $t=0$, and since there’s an even number of slices, we know that there will be another at $t=\pi$. By symmetry, we only need to compute two more cuts. Setting $t_1=0$ produces $t_2\approx 0.77$, and working backwards from the other cut, setting $t_2=\pi$ yields $t_1\approx 1.70$. The resulting pie slices look something like this:
 
If we relax the requirement that all of the cuts radiate from a common point, then there are many more ways to divvy up the pie.
A: Here's an answer to a related problem. 
If it's an elliptical pie and the cuts start at a focus then Kepler's equal area theorem provides the answer. Just send a planet around the edge of the dish and time the orbit.
