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 $\sqrt{k+r_1^2 +r_2^2-\frac{z^2}{4}}$.
Best Answer
$Pow(P, \omega_1) = d(P, O_1)^2 - r_1 ^2$
$Pow(P, \omega_1) = d(P, O_2)^2 - r_2 ^2$
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?