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

• What have you tried? Do you know the equation for the power of a point $P = D^2 - r^2$? Mar 25, 2020 at 17:58
• @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. Mar 25, 2020 at 19:33
• 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. Mar 25, 2020 at 22:12
• @CalvinLin, I wrote it up in the post. Mar 26, 2020 at 6:44
• Nicely done! That's what I got too. Nice relating it to Apollonius circles. Mar 26, 2020 at 13:43

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

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.

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

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