<s>[INST] 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.
 [/INST] 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.
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