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\begin{document}

\title{Project:\ The space of circles.}
\maketitle

\begin{enumerate}
\item We can map $\mathbb{R}^{4}$ to a set of curves in the plane by the map
\[
\mathcal{C}:\left(  a,p,q,r\right)  \rightarrow\left\{  \left(  x,y\right)
\in\mathbb{R}^{2}:a\left(  x^{2}+y^{2}\right)  -2px-2qy+r=0\right\}  .
\]
We may then take a subset of the image curves to be circles. Let $\sim$ denote
the equivalence relation $\left(  a,p,q,r\right)  \sim\left(  \lambda
a,\lambda p,\lambda q,\lambda r\right)  $ for any $\lambda\neq0.$ Then the map
above induces a map from $\mathbb{R}^{4}/\sim$. We consider this set the set
of generalized circles in the plane. Find a subset of the domain so that the
map is a homeomorphism to the space of (nondegenerate) circles in the plane.
(Hint: define the topology on the space of curves so that this is easy.) Find
a formula for the radius and center of $\mathcal{C}\left(  a,p,q,r\right)  $
if it represents a circle.

\item We can define a bilinear \textquotedblleft dot\textquotedblright%
\ product on $\mathbb{R}^{4}$ in the following way%
\[
\left(  a,p,q,r\right)  \cdot\left(  a^{\prime},p^{\prime},q^{\prime
},r^{\prime}\right)  =\frac{1}{2}\left(  2pp^{\prime}+2qq^{\prime}-ra^{\prime
}-ar^{\prime}\right)  .
\]
Recall that given an inner product $\left\langle \cdot,\cdot\right\rangle $
and a symmetric bilinear product $b\left(  \cdot,\cdot\right)  $ on a vector
space $V,$ a number $\lambda$ is said to be an eigenvector of $b$ with respect
to $\left\langle \cdot,\cdot\right\rangle $ if there exists a vector $v\in V$
(said to be the corresponding eigenvector) such that $b\left(  v,w\right)
=\lambda\left\langle v,w\right\rangle $ for all $w\in V.$ (Note that this can
also be formulated in another way. The inner product gives an isomorphism
between $V$ and $V^{\ast}$ given by $\phi:v\rightarrow\left\langle
v,\cdot\right\rangle .$ We also see that $b\left(  v,\cdot\right)  \in
V^{\ast}$ and then map this transformation to $V$ using the inner product.
This gives a linear map. An eigenvalue of the bilinear product is the
eigenvalue of that linear transformation.) What are the eigenvalues of this
bilinear product with respect to the standard inner product on $\mathbb{R}%
^{4}$? Show that if $\mathcal{C}\left(  a,p,q,r\right)  $ and $\mathcal{C}%
\left(  a^{\prime},p^{\prime},q^{\prime},r^{\prime}\right)  $ are intersecting
circles,
\[
\left(  a,p,q,r\right)  \cdot\left(  a^{\prime},p^{\prime},q^{\prime
},r^{\prime}\right)  =-aa^{\prime}RR^{\prime}\cos\phi
\]
where $R$ and $R^{\prime}$ are the radii of $\mathcal{C}\left(
a,p,q,r\right)  $ and $\mathcal{C}\left(  a^{\prime},p^{\prime},q^{\prime
},r^{\prime}\right)  $ respectively and $\phi$ is the angle between the two
circles. Give a condition for two circles to be exteriorly tangent. Describe
the vectors with zero length, positive length, and negative length.

\item Prove Descartes' circle theorem: If four circles in the plane with radii
$R_{1},R_{2},R_{3},R_{4}$ are mutually tangent and their interiors are
disjoint, then
\[
\left(  \frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}%
}\right)  ^{2}=2\left(  \frac{1}{R_{1}^{2}}+\frac{1}{R_{2}^{2}}+\frac{1}%
{R_{3}^{2}}+\frac{1}{R_{4}^{2}}\right)  .
\]
Hint: Consider the generalized circle $\Theta=\left(  0,0,0,1\right)  $ which
must be a linear combination of the four circles since the vector space is
4-dimensional. Then take dot products of the linear dependence relation with
each of the circles and solve a linear system.

\item Describe the closure of the set of circles both in $\mathbb{R}^{4}/\sim$
and geometrically. (Hint: it should include lines.) Use this to generalize
Descartes' circle theorem to the case when one circle is a line.

\item Consider the generalization to $\left(  n-1\right)  $-dimensional
spheres in $\mathbb{R}^{n}.$
\end{enumerate}


\end{document}