Is there any intuitive meaning to $\beta^T A \beta=1$? In a statistical paper, I found that $\beta^T A \beta=1$, where $\beta$ and $A$ are a vector and matrix of constants, respectively.
In the paper, the author utilizes only the $\beta$ that satisfies $\beta^T A \beta=1$, so I guess that the equation stating "a quadratic form is one" has something special meaning.
Is there any special meaning of that? In my rough idea, the quadratic form looks like theequation of a circle.
(I intentionally omitted the exact context, because they are redundant for my question)

Summary: The expression $\beta^\prime A\beta$ generalizes the usual squared Euclidean distance.  The set of $\beta$ for which $\beta^\prime A \beta = 1$ is often a sphere, when considered from the correct perspective; and generally it is always a linear transformation of a cylinder defined over a sphere or hyperboloid.  (A "sphere" in two dimensions is, of course, a circle, as mentioned in the question.)  These form the set of all possible quadratic hypersurfaces centered at the origin of a vector space.

The function
$$Q_A: \beta\to\beta^\prime A \beta$$
is a quadratic form.  By definition, a quadratic form on a vector space $V$ over a field of scalars $R$ is an $R$-valued function $Q$ on $V$ for which
$$Q(\lambda x) = \lambda ^2 Q(x)$$
for all vectors $x$ and scalars $\lambda$ and the function
$$(x,y) \to Q(x+y) - Q(x) - Q(y)$$
(defined on ordered pairs of vectors) is linear in both variables [Serre Chapter IV Definition 1].  The properties of matrix multiplication readily imply both of these properties of $Q_A.$
The second requirement is needed only when $1+1=0$ in the field $R.$  Because that rarely interests statisticians, I will assume from now on that this is not the case.
BTW, the first condition immediately shows there's nothing special about the "$1$" in the equation $Q_A(\beta)=1$ except that it is positive, because all level sets of positive value are just rescaled versions of this one.
The first reason for placing the question in this general setting is to point out that different matrices $A$ and $A^\prime$ can define the same quadratic form.  The reason is evident when you write out the product in $Q_A.$  This is a combination of products of components of $\beta.$  For indices $i\ne j,$ the coefficient of $\beta_i\beta_j$ is $A_{ij} + A_{ji},$ which is the $(i,j)$ coefficient of the symmetric matrix $A + A^\prime.$  For example, these matrices all define the same quadratic form:
$$\pmatrix{0&-2\\0&0},\quad \pmatrix{0&0\\-2&0},\quad \pmatrix{0&1\\-3&0},\quad \pmatrix{0&-1\\-1&0},$$
because in all cases $Q_A((\beta_1,\beta_2)^\prime) = -2\beta_1\beta_2.$  Notice that in this example the set of vectors $\beta\in\mathbb{R}^2$ for which $Q(\beta) = 1$ comprises both branches of the hyperbola $\beta_1\beta_2 = -1/2.$
This should make it clear that

Whenever $A$ and $B$ are square matrices of the same size and $A+A^\prime = B+B^\prime,$ the quadratic forms $Q_A$ and $Q_B$ are the same.

Consequently, when the function $Q_A$ is the object of interest, (1) the details of the matrix $A$ are irrelevant and (2) we may always assume $A = A^\prime$ is symmetric.
The second reason is that linear algebra offers a host of subtly related matrix decompositions.  Recognizing that we're really dealing with a quadratic form strips away inessential characteristics of the problem, allowing us to focus on the relevant theorems and making it easier to see what $A$ is doing.  In particular, the fundamental theorem about quadratic forms on finite-dimensional Real vector spaces is that they can always be diagonalized with a suitable orthogonal matrix.  With a little post-processing (see the inv function in the code at the end of this post), this implies there exists a matrix $P$ (of the same size as $A$) for which
$$Q_A(\beta) = (P\beta)^\prime J (P\beta)$$
where $J$ is a diagonal matrix with only the values $\pm 1$ and $0$ on the diagonal.  We can also arrange, if we wish, for the first $k$ values to be $1,$ the next $m$ values to $-1,$ and the remaining values to be $0.$ Sylvester's Law of Inertia asserts that this is possible and, no matter how it's done -- there usually are many matrices $P$ that diagonalize a form in this way -- you always obtain the same values of $k$ and $m,$ known as the "signature" of $Q_A.$
The columns of the matrix $P$ form a basis of $V.$  Any basis establishes a way to name each vector in $V$ uniquely (as a linear combination of the basis).  In this basis, the formula for $Q_A$ is particularly simple.  Writing $\alpha = P\beta$ for such a linear combination, Sylvester's Law says

$$Q_A(\beta) = \alpha_1^2 + \cdots + \alpha_k^2 - (\alpha_{k+1}^2 + \cdots \alpha_{k+m}^2).$$

The set of $\alpha$ for which $Q(\beta) = 1$ thereby describes a spheroid or hyperboloid in $V$.  In most statistical applications, $Q_A$ is positive definite.  This means $k=n$ (the dimension of $V$).  Equivalently, $J$ is the identity matrix if and only if $Q_A$ is positive-definite.  (We say $Q_A$ is positive semidefinite when $J$ has no values of $-1$ but might have some zeros on the diagonal.)
You should recognize this as the usual formula for squared Euclidean distance: it embodies the Pythagorean Theorem.
In other words,

when $Q_A$ is positive definite, $Q_A$ is the square of the usual Euclidean length (from the perspective of a suitable basis depending on $A$); and the function $(x,y)\to Q(x+y) - Q(x) - Q(y)$ is four times the usual dot product in that basis.

The level set $Q_A(\beta)=1$ in this case corresponds to the unit sphere, $\alpha_1^2 + \alpha_2^2 + \cdots + \alpha_n^2 = 1$ where again $\alpha = P\beta.$
In the original basis of $V$ (used for expressing $\beta$), the unit sphere has been distorted by the mapping $P^{-1}.$  It will be an ellipsoid.
To continue the previous example, the matrix $P$ can be taken to be
$$P = \frac{1}{\sqrt{2}}\pmatrix{-1&-1\\1&-1}.$$
You can readily compute
$$P^\prime\ (A + A^\prime)/2\ P = \pmatrix{1 & 0 \\ 0 & -1},$$
whose signature is $(1,-1).$  That is,
$$\begin{aligned}
-2\beta_1\beta_2 &= \beta^\prime A \beta = (P\beta)^\prime \pmatrix{1 & 0 \\ 0 & -1} (P\beta) = (P\beta)_1^2 - (P\beta)_2^2 \\
&= \frac{1}{2}(-\beta_1+\beta_2)^2 - \frac{1}{2}(-\beta_1-\beta_2)^2
\end{aligned}$$
for all vectors $\beta = (\beta_1,\beta_2)^\prime.$
To analyze an arbitrary $Q_A,$ then, you will want to find the matrix $P$ that diagonalizes the matrix $A+A^\prime,$ but even before you compute $P$ you know you are working with ordinary Euclidean geometry (when $A+A^\prime$ is positive definite) or, at worst, in a pseudo-Riemannian metric, of which the Minkowski space of the theory of Special Relativity is the most notable example (whose metric has signature $(3,-1)$).

As a demonstration of the computational practicality of this way of looking at quadratic forms, here is R code to diagonalize any square matrix. It generates a random square matrix A, diagonalizes it, and exhibits its diagonal form.
# Generate a random square matrix.
# set.seed(17)
n <- 5
A <- matrix(rnorm(n^2), n)

# Compute the symmetric representative of its quadratic form.
A <- (A + t(A))/2

# Diagonalize it.
obj <- eigen(A)
inv <- function(x) ifelse(x == 0, 0, 1/x)
P <- t(t(obj$vectors) * inv(sqrt(abs(obj$values))))

# Check the diagonalization.
zapsmall(t(P) %*% A %*% P)

When I ran it after setting the seed to 17 it output
     [,1] [,2] [,3] [,4] [,5]
[1,]    1    0    0    0    0
[2,]    0    1    0    0    0
[3,]    0    0    1    0    0
[4,]    0    0    0   -1    0
[5,]    0    0    0    0   -1

The eigen function in R will cause the output to have first the 1's on the diagonal, then any zeros, then the -1's at the end.
Reference
Serre, J-S. 1973.  A Course in Arithmetic.  Springer-Verlag.
