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!

Pingback: Parabolas of Apollonius? | Girls' Angle

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.

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

Later, Thornton received word from the DMV that his license to drive was suspended.

They consider to keep drivers out of prison and help them

from dropping their driving licenses. The attorneys at are renowned for aggressively defending the

DUI cases of their clients. If FMLA doesn’t cover your sickness, you can get fired. Typically, you will apply to law school during the final year of college.