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 $\sqrt{k+r_1^2 +r_2^2-\frac{z^2}{4}}$.

  • $\begingroup$ What have you tried? Do you know the equation for the power of a point $P = D^2 - r^2$? $\endgroup$
    – Calvin Lin
    Commented Mar 25, 2020 at 17:58
  • $\begingroup$ @CalvinLin For 2 orthogonal circles with centers $O_1, O_2$, the locus is a circle with diameter $O_1O_2$. Of course I know all about the radical axis and power of a point theorems. I tried applying cosine law in triangle $PO_1O_2$ where $P$ is the locus point, which resulted in something like: $d_1d_2\cos\theta$ is constant. $\endgroup$
    – AvidSeeker
    Commented Mar 25, 2020 at 19:33
  • $\begingroup$ Hm, can you write that out? That's all that I did to arrive at a reasonable characterization (which includes your special case). I will undelete my answer when you show your work. $\endgroup$
    – Calvin Lin
    Commented Mar 25, 2020 at 22:12
  • $\begingroup$ @CalvinLin, I wrote it up in the post. $\endgroup$
    – AvidSeeker
    Commented Mar 26, 2020 at 6:44
  • $\begingroup$ Nicely done! That's what I got too. Nice relating it to Apollonius circles. $\endgroup$
    – Calvin Lin
    Commented Mar 26, 2020 at 13:43

2 Answers 2


$Pow(P, \omega_1) = d(P, O_1)^2 - r_1 ^2$
$Pow(P, \omega_1) = d(P, O_2)^2 - r_2 ^2$

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"

  • $\begingroup$ Hi, thanks for the reply, but I didn't see the answer in the link you provided. $\endgroup$
    – AvidSeeker
    Commented Mar 25, 2020 at 19:44
  • $\begingroup$ -1 OP is not asking for "$Pow (P, \omega_1 ) = Pow (P, \omega_2)$, which is the equation for the radical axis. $\endgroup$
    – Calvin Lin
    Commented Mar 25, 2020 at 22:06

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