It is not very clear to me what you mean by "intersection of Hilbert Class Fields [...] is discussed". The theory of Complex Multiplication (see, for instance, Serre's short note in Cassels and Frohlich, or Silverman's _Advanced Topics in the Arithmetic of Elliptic Curves_, or directly the bible from Shimura, _Introduction to the Arithmetic Theory of Automorphic Functions_) tells you that there is an explicit way, given an imaginary quadratic field $K=\mathbb{Q}(\sqrt{d})$ for constructing not only its Hilbert class field, but all ray class fields of different conductors. For instance, for the Hilbert class field, you can first create the elliptic curve $\mathbb{C}/\mathcal{O}_K$: it is an elliptic curve with complex multiplication by $\mathcal{O}_K$. Then, the theory will tell you that the smallest extension of $\mathbb{Q}$ containing $\sqrt{d}$ over which this curve is defined is the Hilbert Class Field of $K$. Just to be convinced that what I say is believable (beside being true, for which you might look at the references), observe that there is an elliptic curve with CM by $\mathbb{Z}[i]$ which admits a Weierstrass equation
$$
y^2=x^3+x\;.
$$
This is defined over $\mathbb{Q}$, so the smallest field of definition which contains $\sqrt{-1}$ is $\mathbb{Q}(i)$, which is indeed its own Hilbert class field.

In more concrete terms, if you are given a squarefree $d,0$ and you want to discover the Hilbert Class Field of $K=\mathbb{Q}(\sqrt{d})$, it is the field $K(j(\sqrt{d})$, where $j$ is the modular function
$$
j(q)=\frac{1}{q}+744+196884q+\dots
$$
whose definition you'll find in the above references, or at
http://en.wikipedia.org/wiki/J-invariant

In your case, life is easier: you only want to be sure that if $d\equiv 1\pmod{4}$ is negative and squarefree, then $K(i)/K$ is unramified (being abelian, this would force it to lie inside the Hilbert Class Field). For this, it is enough to observe that there is a diagram
$$
\xymatrix{
&K(i)&\\
K&\mathbb{Q}(i)&F:=\mathbb{Q}(\sqrt{-d})\\
&\mathbb{Q}&
}$$
Now pick a prime $\ell$ dividing $d$: it is necessarily odd. Its ramification degree is $2$ both in $K$ and in $F$, while it is unramified in $\mathbb{Q}(i)$. Therefore it needs be unramified in $K(i)/K$, since ramification degrees are multiplicative in towers of extensions. Similarly, $2$ is unramified in $F/\mathbb{Q}$ and cannot ramify in $K(i)/K$. For what concerns the infinite primes, observe that $K$ is already totally complex. Done!