A heart shape is constructed using two identical circles with radius $r$. A line is drawn from point $T$ to $C$. There, two tangents are constructed, one to each circle, through $AC$ and $CB$. (The heart is supposed to be symmetrical, I apologize for the rough sketch)
Presuming that the coordinates of points $A,B,C,T$ and length of $r$ as well as the area $A_\text{Heart}$ are known, I would like to calculate the point $S$, the centroid of the heart.
Now, if the heart is "tall" enough, calculating the centroid would be easy. As the heart is perfectly symmetrical, $S_x = C_x = T_x$. The centroid lies in the center of the line that seperates the heart into two areas of equal size ($\frac{1}{2} A_\text{Heart}$).
$$\frac{1}{2} \overline{SR} \overline{SC} = \frac{1}{2} A_\text{Heart} $$
Through basic trigonometry, the knowledge of the angle $\alpha$ gives us
$$ \overline{SR} = 2 \tan(\alpha) \overline{SC} $$
$$ \overline{SC} = \sqrt{ \frac{A_\text{Heart}}{2 \tan(\alpha)}}$$
This leads to the final formula:
$$S(S_x|S_y) = \left(C_x \mid C_y - \sqrt{ \frac{A_\text{Heart}}{2 \tan(\alpha)}}\right)$$
This approach assumes that the area of the triangle $ABC$ is greater than the rest of the heart, meaning that the centroid must lie within the triangle. If the heart were of shorter height, the centroid will inevitably lie outside of the triangle.
I'm open to any ideas.
If the centroid is above the triangle, you want to compute the height that has half the area above it. The centroid is clearly below point $T$ as $T$ splits the area of the circles in half and the triangle will lower the centroid. You want the area of the strip from $T$ down to $S$ to be $\frac 12A_{heart}-\pi r^2$ where the outer edge of the strip follows the circle. Put the origin at $T$ with the $x$ axis directed toward $S$. The area of the strip is $2\int_0^{|TS|} r+\sqrt{r^2-x^2}dr=\frac 12A_{heart}-\pi r^2$ Solve for $x$