Locus of points satisfying a given condition. Find the locus of the points P in the plane of an equilateral triangle ABC for which the triangle formed with PA, PB, and PC has constant area.
I have used the Heron's formula but no avail ( I already suspected that this problem could not be easily solved by this formula but I had no other idea since the only thing I know about the triangle is its sides PA PB PC ). I have no idea how to proceed. So any help will be appreciated.
Update : I have seen that futurologist says that the locus will be any circle centred at the circumcentre of ΔABC. But can anyone say why this is so?
 A: Let $PA=a$, $PB=b$ and $PC=c$ be the distances of point $P$ from the vertices of equilateral triangle $ABC$. By Heron's formula, if $S$ is the area of the triangle of sides $a$, $b$ and $c$ we have
$$
\tag{1}
16S^2=(a+b+c)(a+b-c)(a-b+c)(-a+b+c)=4b^2c^2-(b^2+c^2-a^2)^2.
$$
Choose now the coordinates of $A$, $B$ and $C$ so that the center of $ABC$ is at the origin:
$$
A=(r,0),\quad B=\left(-{1\over2}r,{\sqrt3\over2}r\right),
\quad C=\left(-{1\over2}r,-{\sqrt3\over2}r\right),
$$
where $r$ is the radius of $ABC$. If $P=(x,y)$ we then have
$$
a^2=(x-r)^2+y^2,\quad
b^2=\left(x+{1\over2}r\right)^2+\left(y-{\sqrt3\over2}r\right)^2,\quad
c^2=\left(x+{1\over2}r\right)^2+\left(y+{\sqrt3\over2}r\right)^2.
$$
Substituting that into $(1)$ one gets:
$$
16S^2=3(x^2+y^2-r^2)^2,
\quad\hbox{that is:}\quad
x^2+y^2=r^2\pm{4\over\sqrt3}S.
$$
The latter is the equation of two circles centered at the origin, which are then the required locus. They are both real if $r^2>{4\over\sqrt3}S$, that is if $S<{1\over3}S_{ABC}$, and have radii
$\sqrt{r^2\pm{4\over\sqrt3}S}$. If instead $S>{1\over3}S_{ABC}$, then only one of them is real.
EDIT.
The diagram below shows that for $S<{1\over3}S_{ABC}$ the locus is actually formed by two circles. The upper triangle on the right has sides congruent to $PA$, $PB$ and $PC$, the lower triangle has sides congruent to $QA$, $QB$ and $QC$. Both triangles have the same area.

A: Area of a triangle $\Delta $ with sides $(a,b,c)$ by simplified Heron/Brahmagupta formula written in the required form:
$$ 16 \Delta^2  = 2(a^2b^2 + b^2c^2+c^2a^2) -(a^4+b^4+c^4) \tag{1}$$
We take an equilateral triangle of circum-radius $2R$ with vertices (taken double for convenience)
$$ (2R,0), (-R,\pm \sqrt {3} R) \tag{2}$$
and compute sides of triangle of sides as three distances between the above vertices and variable point $ P (x,y) $ for required locus of same or constant area $\Delta$ enclosed. 
$$ a^2 = (x-R)^2 + y^2;\, b^2 =(x+R)^2 +( y-\sqrt 3 R )^2;\,c^2 =(x+R)^2 +( y+   \sqrt 3 R )^2 \, ; \tag{3}$$
Plug into (1) and simplify algebraically, (CAS aided) ..
$$  |\Delta| = \dfrac{\sqrt {3}}{4} [(x^2 +y^2) - (2 R)^2]  \tag{4} $$
$$  |\Delta| = \dfrac{\sqrt {3}}{4} [(x^2 +y^2) - (r_c)^2]  \tag{5} $$
which are all Circle loci centered at origin and circum-radius $2R=r_c$
It is noticed that area of the triangle vanishes(becomes zero) when $P$ is taken as one vertex because sides $ ( \sqrt {3} R, \sqrt {3}R,0) $ cannot enclose any area. We have positive area for point $P$ outside the circum-circle as shown and negative when inside and zero on the circum-circle. 
This can be verified with a short trig check for the three rays inside the circle, as $ (b+c-a) $ is one factor of area.
$$ b+c-a = PD\, [\cos(\pi/3 +u) + \cos(\pi/3 -u)- \cos u ] = PD \, ( 2 \cos \pi/3 \cos u - \cos u ) =0  \tag {6} $$
Area cannot be enclosed if on re-arrangement $a,b,c$ lie along a straight line, as it happens here. (or if $ a < b+ c $ which does not arise here).   $60^0$ angles marked in red.

$$ 4R \Delta = abc$$
In fact by virtue of the above there is no need for any further calculation, we have constant areas $A$ for magnified circium radii !! due to geometric similarity...
Finally, if side length of each side of equilateral triangle $  ABC= L$ is given and the area $ |\Delta| =A $ of a triangle  connecting sides to vertex $P$ is also given as $ A,$ so that circum-radius is $ r_c= \dfrac{L}{\sqrt3},$ then by virtue of (4) we have for  positive Area Regime (Faint Blue and brown shaded annular rings):
$$ \boxed{ r_{locus} = \sqrt{ \frac{L^2}{3} +\frac{4A}{\sqrt3}  } } \tag{6} $$

EDIT1:
Negative area Regime ( Dark and light Blue annular rings) has meaning/sense here as at a certain $ r_{locus} $ area is zero.  The length and area cannot be arbitrarily/independently given beyond a certain limit in this regime. 
The graph shows positive regime above x-axis and negative regime below x-axis as a parabola  of quadratic equation solution for both signed  areas $A$, continuously with respect to changing locus radius.
$$ \boxed{ r_{locus} = \sqrt{ \frac{L^2}{3} -\frac{4A}{\sqrt3}  } } \tag{7} $$
A plot of Equation (7) shows that computed area of triangle with sides $ PA,PB,PC, $
$$ |A| \gt \sqrt3 L^2/12 $$
is not permissible for real $\, r_{locus}. $ 
