Circles of Apollonius… and magnetism!

One way to simplify things is to pick one of the circles of Apollonius and make its center be the origin and its radius be $1$. Then, points $A$ and $B$ are on the $x$-axis, at a distance $1/c$ and $c$ from the origin, respectively, for some positive real number $c>1$. Then, pick a point $P=(x,y)$ on the circle, and let $v$ and $w$ be the magnetic field vectors at that point coming from the wires at $A$ and $B$ respectively.

We want to show that the purple vector, $w+v$, is a tangent vector to the circle at point $P$. It’s tangent if it’s perpendicular to $OP$, so in particular, we want to show that the ratio of its $y$-coordinate to its $x$-coordinate is $-x/y$.

But this isn’t hard to show given what we know about $v$ and $w$. These are the magnetic field vectors, so they are perpendicular to $AP$ and $BP$ respectively. This gives us the equations $v\cdot (x-1/c,y)=0$ and $w\cdot (x-c,y)=0$, or equivalently,
$$v_1x+v_2y-v_1/c=0$$ and $$w_1x+w_2y-w_1c=0.$$

Adding these equations and dividing by $v_1+w_1$, we find
$$\frac{v_2+w_2}{v_1+w_1}=\frac{v_1/c+cw_1}{y(v_1+w_1)}-x/y.$$

But we want to show that the right hand side is simply $-x/y$, so it suffices to show that $v_1=-c^2w_1$.

Now, because the strength of the magnetic field is inversely proportional to the distance from the wire, and the circle is the locus of points with a fixed ratio to $A$ and $B$, we also have that
$$|v|/|w|=PB/PA=QB/QA=(c-1)/(1-1/c)=c$$ where $Q$ is the intersection point of $AB$ with the circle. So, $$v_1^2+v_2^2=c^2(w_1^2+w_2^2).$$

Since we are only interested in the ratio of $v_1$ to $w_1$, let us set $w_1=1$ for simplicity, and try to show that $v_1=-c^2$. If $w_1=1$, then $w_2=(c-x)/y$ from the equation above, and so $w_1^2+w_2^2=(1+c^2-2cx)/y^2$. It follows that
$$v_1^2+v_2^2=c^2(1-c^2-2cx)/y^2,$$ and we had $$v_1x+v_2y-v_1/c=0$$ from before. The values $v_1=-c^2$ and $v_2=(c^2x-c)/y$ satisfy both equations, and so $v_1=-c^2w_1$ as desired.

So there we have it! The circles of Apollonius appear as the magnetic field lines between two parallel, oppositely-oriented, current-carrying wires. Beautiful!

4 thoughts on “Circles of Apollonius… and magnetism!

  1. Pingback: Parabolas of Apollonius? | Girls' Angle

  2. Another pretty occurrence of circles of Apollonius is this:

    Consider the disk model for hyperbolic geometry. The hyperbolic plane corresponds to an open disk whose (Euclidean) boundary can be called a circle at infinity. Take two points on the circle at infinity. Take the Euclidean circle of Apollonius corresponding to a fixed distance ratio (or line, if the ratio is 1). Then its intersection with the open disk is a hyperbolic line!
    If you do this in d dimensions instead of 2, the ‘(d-1)-spheres of Apollonius’ corresponding to a fixed distance ratio (or (d-1)-plane, if the ratio is 1) will again give a hyperbolic (d-1)-plane. But in this setting, I can’t promise the connection with magnetic fields carries over.

  3. Pingback: “Parabolas of Apollonius?” Spoiler | Girls' Angle

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