Lets say we have two circles call them $O_{1}$ and $O_{2}$. Let $a_{1}$ and $a_{2}$ be the arcs of the circles. Then when it comes to two circles, three cases arise. They intersect at two points, they intersect at one point, or they intersect at no points. Let $\omega_{a_{1}}(P)$ denote the power of a point $P$ from circle $O_{1}$. Let $\omega_{a_{2}}(P)$ denote the power of a point $P$ from circle $O_{2}$. If we define $\{P | \omega_{a_{1}}(P) = \omega_{a_{2}}(P)\}$ then what would this be equal to for all three cases?
My thoughts : If the circles intersect at two points then $P$ would just be on the line connecting the two points right? That would be where they are equal at? If that were true, how would I write that? i.e. $\{P | \omega_{a_{1}}(P) = \omega_{a_{2}}(P)\} =$?
For the case when the circles intersect at one point, $P$ is just on the tangent line both of these circles share? Again how would I write this?
The last case I am a little unsure. Maybe one can draw a line between the circles s.t. the distance from the center of each circle to the line is the same and then $P$ would be on that line?
The power of a point $p$ relative to a given circle is $$ |p-c|^2-r^2\tag{1} $$ where $c$ is the center of the circle and $r$ its radius. Thus, given two circles $(c_1,r_1)$ and $(c_2,r_2)$ the power of $p$ relative to both circles is equal when $$ \begin{align} |p|^2-2p\cdot c_1+|c_1|^2-r_1^2&=|p|^2-2p\cdot c_2+|c_2|^2-r_2^2\\ 2p\cdot(c_2-c_1)&=|c_2|^2-|c_1|^2+r_1^2-r_2^2\tag{2} \end{align} $$ The locus of $p$ is a line perpendicular to $c_2-c_1$; that is, the line between the centers. Since the power of points on each circle is $0$, the line must pass through the points of intersection (or tangency).
For circles that do not intersect, the locus passes through the mid points of the common tangents:
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