Find the area enclosed by $\sqrt{(x-2)^2+(y-3)^2} + 2\sqrt{(x-3)^2+(y-1)^2} = 4$ 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.
 A: 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

A: Here's the answer using polar coordinates. It turns out that the implicit curve is quadratic in $r$ in polar coordinates which means $r$ can be solved for in terms of the angle $t$ using the quadratic formula. Finally, using $A=\frac{1}{2} \int_0^{2 \pi} r(t)^2 dt$ allows for the area to be computed. 
For simplicity, take $x_1=y_1=0$ since we can place the origin anywhere. We start with
$$a \sqrt{x^2+y^2}+\sqrt{(x-x_2)^2+(y-y_2)^2}=k.$$
Notice I swapped where the $a$ occurs. It will make some slightly algebra nicer. Substituting $x=r \cos t$ and $y= r \sin t$ gives
$$\sqrt{(r\cos t-x_2)^2+(r \sin t-y_2)^2}=k-ar$$
and after squaring
$$r^2 - r(2 x_1 \cos t +2 y_2 \sin t+1)+(x_1^2+y_2^2)= k^2-2kar + a^2r^2.$$
With dependencies on $t$, this function has the form
$$r^2-r(A\cos t + B \sin t + C) + D = 0$$
where $A = \frac{2x_1}{1-a^2}$, $B = \frac{2y_2}{1-a^2}$, $C = \frac{1+2k}{1-a^2}$, $D = \frac{x_1^2+y_1^2-k^2}{1-a^2}$. Given our initial parameters, perhaps there are bounds on $A,B,C$ and $D$. I don't care. I'll take the above as the new definition of our curve and find the area in terms of the new parameters making assumptions as I go.
Using the quadratic formula yields
$$ r = \frac{1}{2}\left[(A\cos t + B \sin t + C)\pm \sqrt{(A\cos t + B \sin t + C)^2-4D} \right].$$
We have the form $r(t)=p(t) \pm \sqrt{q(t)}$ for $r$. Assume parameters are chosen so that $q(t)>0$. If we $D<0$ we have one branch, otherwise two curves satisfy the original equation. Then the area of the region enclosed is given by
$$A = \frac{1}{2}\int_0^{2\pi} r(t)^2 dt = T_1 + T_2 + T_3,$$
with $T_1 = \frac{1}{2}\int_0^{2\pi} 2 p(t)^2 dt$, $T_3 = \pm\frac{1}{2}\int_0^{2\pi} p(t) \sqrt{q(t)} dt$, and $T_3 = \frac{1}{2} \int_0^{2\pi} q(t) dt$.
More explicitly, this is 
$$T_1 = \frac{1}{8} \int_0^{2\pi} (A \cos t+B \sin t+C)^2 = \frac{1}{8}(\frac{2 \pi A^2}{2} + \frac{2 \pi B^2 }{2} + 2\pi C) = \frac{\pi}{8}(A^2+B^2+C)$$
$$T_2 = \pm \frac{1}{4} \int_0^{2\pi} (A\cos t + B \sin t + C) \sqrt{(A\cos t + B \sin t + C)^2-4D} \; dt$$
$$T_3 = \frac{1}{8} \int_0^{2\pi} \left[(A\cos t + B \sin t + C)^2-4D \right] dt = \frac{\pi}{8}(A^2+B^2+C-4D)$$
As seen, the first and third terms are easily computable. The second requires work. Because $A \cos t + B \sin t = \alpha \sin(t+\delta)$ for $\alpha = \sqrt{A^2+B^2}$ and $\delta$ depending on $A$ and $B$, we can substitute this in the expression for $T_2$ and do a change of variables to get
$$T_2 = \pm\frac{1}{4} \int_0^{2\pi} (\sqrt{A^2+B^2} \sin t + C) \sqrt{(\sqrt{A^2+B^2} \sin t + C)^2-4D} \; dt.$$
I highly doubt the antiderivative of the above can be expressed in terms of elementary functions. Perhaps someone can work this into an expression involving elliptic integrals. 
If I get to it, I'll work out the degenerate case of an ellipse. In that situation, my guess is that the $T_2$ term vanishes. I've almost certainly made an error or two--please let me know if I did. However, the overall approach is sound.
A: from geometry point, the equation $\sqrt{(x-2)^2+(y-3)^2} + 2\sqrt{(x-3)^2+(y-1)^2} = 4$ is equal to:
$\sqrt{x^2+y^2} + 2\sqrt{(x-\sqrt{5})^2+y^2} = 4$ or $\sqrt{(x-\sqrt{5})^2+y^2} + 2\sqrt{x^2+y^2} = 4$ which is symmetry to $x$ axis  
so for first one:
$y = \dfrac{ \sqrt{-9 x^2+24 \sqrt{5} x-16 \sqrt{6 \sqrt{5} x+1}+20}}{3}$ for the half area.
but the integration is difficult. I have no idea except the numeric method.
edit: the area is:
$2\int_{x_1}^{x_2}ydx=3.2,x_1=2 (\sqrt{5}-2),x_2=\dfrac{2 (\sqrt{5}+2)}{3}$
and for general a,b,if it is not rational,it is complex. if it is rational, then we can simplify in same idea.
