Sum of points' powers to circles is constant Given 2 circles $\omega_1, \omega_2$, find the locus of all points $P$ such that $\mathcal{P}ow(P, \omega_1) + \mathcal{P}ow(P, \omega_2) = k$ (i.e: sum of powers of point $P$ with respect to the two circles $\omega_1, \omega_2$ is constant).

UPDATE: I found out two solutions:

1. Analytical geometry approach:

let $O_1=(0, 0), r_1, O_2=(z, 0), r_2$ be the centers and radiuses of $\omega_1, \omega_2$ respectively. Denote by $d_1, d_2$ to be the distances between $P$ and $O_1, O_2$. Let $P=(x, y)$. By the definition of power of a point: $d_1^2 -r_1^2+d_2^2-r_2^2=k\Leftrightarrow d_1^2 +d_2^2 =k+r_1^2 +r_2^2$
$$
\Leftrightarrow x^2 + y^2 + (z-x)^2 +y^2=k+r_1^2+r_2^2\\
\Leftrightarrow (x-\frac{z}{2})^2+y^2=\frac{k+r_1^2 +r_2^2-z^2}{2} +\frac{z^2}{4}
$$

Which is a circle centered at the midpoint of $O_1, O_2$

2. Euclidean geometry: using Apollonius theorem:

Consider the midpoint $M$ of $O_1O_2$, by applying Apollonius theorem in $\triangle PO_1O_2$: $d_1^2 +d_2^2 = PM^2 +\frac{z^2}{4}$ This means that $PM$ is constant. Therefore, $P$ is on the circle centered at $M$ with radius $\sqrt{k+r_1^2 +r_2^2-\frac{z^2}{4}}$.
$Pow(P, \omega_1) = d(P, O_1)^2 - r_1 ^2$
$Pow(P, \omega_1) = d(P, O_2)^2 - r_2 ^2$

So, we want to find the set of points $P$ such that 

$d(P, O_1)^2 + d(P, O_2 )^2 = k + r_1^2 + r_2^2$ 

What is the shape of the graph?


   It is a circle, whose center is the midpoint of $O_1O_2$.
 To see this, set $P = (x,y), O_1 = (x_1, y_1), O_2 = (x_2, y_2)$, expand out the equation and simplify.
See here: https://artofproblemsolving.com/wiki/index.php/Radical_axis

I believe this has something to do with something along the lines of "Newton's Lemma"