TITLE: Find the area enclosed by $\sqrt{(x-2)^2+(y-3)^2} + 2\sqrt{(x-3)^2+(y-1)^2} = 4$
QUESTION [23 upvotes]: Question: What is the area of the interior of the simple closed curve described by the equation 
$\sqrt{(x-2)^2+(y-3)^2} + 2\sqrt{(x-3)^2+(y-1)^2} = 4$?
Comments: I came up with this specific problem myself in response to my earlier question, which I don't think was well posed, or at least it was not clear what I was after: to see how to find the area of the interior of a Jordan curve that is described by an implicit function. To see how this area looks like, I uploaded a picture from WolframAlpha:

As you can see, it is quite egg-like. We can generalize; the object could be called a weighted ellipse [edit: usually called a Cartesian oval] with an equation of the form
$\sqrt{(x-x_1)^2+(y-y_1)^2} + a \cdot \sqrt{(x-x_2)^2+(y-y_2)^2} = k$
where $(x_1,y_1)$ and $(x_2,y_2)$ are the Cartesian coordinates of the focal points of this weighted ellipse and $a$ is a weight (it equals $1$ in the case of an ordinary ellipse). As a bonus question, I would like to see how to find the area of this object.

REPLY [6 votes]: Simple minded approach.
Take an area of $3500 \times 3500$ pixels (not really) and count the pixels inside the curve.

DEPRECATED

VERY LATE UPDATE (2021).
Take an area of $3^{13} \times 3^{13}$ pixels (not really) and count the pixels inside the curve.
The following method is supposed to work for any convex closed (Jordan) curve that divides the plane.
Start with a grid of $3 \times 3$ squares and refine the grid three times each iteration.
Leave out a square as soon as all four vertices of it are guaranteed to be inside or outside the curve.
Continue (recursively) with the remaining squares, until the maximum grid refinement is reached.
Pictures say more that a thousand words:

Free accompanying (Delphi Pascal) source code available at this place:

MSE publications / references 2021

Output after 13 iterations:
$$
  3.204044 < \mbox{Area} < 3.204057
$$
Which is quite in agreement with (and inferior to) the above answer given by achille hui.
There is also a version available with binary instead of ternary refinement.
In addition, the software has been enhanced with Isoline Refinement.
The meaning of the latter is best explained with a picture ($\color{red}{red}$  line):

With isoline refinement, a much better approximation of achille hui's result is found.
Agreement is now within 12 digits. As a bonus, a value for the length of the perimeter is obtained as well:

Area = 3.20405086915288E+0000
Perimeter = 6.457419618 86746E+0000