How would I cut one washer into two equal-area washers? I'm trying to mill an asymmetric graphite crucible in the shape of a hollow cone (imagine an ice cream cone with the end bitten off).  I require identical "horizontal resistance" at the wide top, as well as the narrow bottom (and throughout) - hence - the walls of my cone need to be fatter at the top, and they taper towards the bottom.  Resistance goes up with length, and down with area - my length (around the circular path) gets longer at the top, so I need more area to keep it constant.  (This is a hobby project, not a job or homework)
Viewed from above, the start of the cone is a disc with a hole in it - e.g. the shape of a big washer with a big hole.
Viewed from below, again, the end (that you bit off) is the shape of a washer, but this time, the hole is much smaller.
To calculate the approximate* resistance that electricity** will encounter when travelling from one side to the other of each washer-shaped area, I decided to draw an imaginary line on this washer such that it equally divides the total area.
*= it enters sideways from a square join-point equal to the outer-inner radius 
**= it is 3v, so no skin effects
Another way to imagine the same problem: if you start with one wide washer with a small hole, how do you calculate the circumference of the line that divides that washer into two equal-area washers? (e.g if you were to cut along this mid-line, you'd get one big washer of the original circumference with a big hole in the middle, and a second smaller washer with the original small hole in the middle, and they would both weigh the same).
I've spent an hour in google with no luck so far (I'm not sure of the terms to search on, but I'm pretty sure Pythagoras isn't helping, and it's dominating the results).
An approach I attempted which didn't work out was as follows:
I reasoned that if the outer radius is A and the hole radius is C, then the midpoint radius can be B.  From there, I reasoned that C is likely to be a radio of A to B somehow, so I defined C as B - n * (A-B), and ended up trying to solve for n, which gave me n=(sqrt(A^2-2B^2)-B)/(A-B) which broke immediately of course when I tried an example because: sqrt(-3).
I'm guessing this is probably harder math than I realized?
Right now, I'm obsessing over how to divide washer-shapes equally.  The real problem (of which I'm fairly sure that answer is part) is how to find the formula to draw concentric expanding washer shapes such that their "midpoint circumferential length" multiplied by their total area is a constant.
Any help/tips/clues are most welcome.  Go easy on me - the stuff I'm trying to do now was drummed into me 40 years ago, plus I know "programmer math" not "symbol math", so to say I'm rusty/unprepared is an understatement...
 A: Interesting geometry problem!
Let’s suppose we have a “washer” (or, a circle with a smaller circle removed from the middle) with outer radius $R$ and inner radius $r$, like this:

We want to solve for $r^*$, the radius of the circle dividing this washer into two smaller equal-area (or equal-mass) washers. 
First, note that the area of an arbitrary washer with inner radius $r$ and outer radius $R$ is given by 
$$\pi R^2-\pi r^2$$
since the washer is the result of removing a circle with area $\pi r^2$ removed from a circle with area $\pi R^2$. Now, we would like to calculate $r^*$ such that the areas of the two “sub-washers” produced is the same, or
$$\pi R^2-\pi {r^*}^2=\pi {r^*}^2-\pi r^2$$
Solving this equation for $r^*$ gives us the solution
$$r^*=\sqrt{\frac{R^2+r^2}{2}}$$
A: If I understand your question correctly, you have two concentric circles in the plane, with radii $r$ and $R$, where $r<R$. And you want to find the radius $x$ of the circle that divides the area between these circles in half.
The outer ring has area $\pi(R^2-x^2)$.
The inner ring has area $\pi(x^2-r^2)$.  
So you want $x$ such that $R^2-x^2=x^2-r^2$. In other words,
$$x=\sqrt{\frac12(R^2+r^2})$$
