<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] 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.
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