Let us first define the two problems:
- Problem 1: \begin{equation} \min_{\beta} ~f_\alpha(\beta):=\frac{1}{2}\Vert y-X\beta\Vert^2 +\alpha\Vert \beta\Vert^2\end{equation}
- Problem 2: \begin{align} \min_{\beta} ~&\frac{1}{2}\Vert y-X\beta\Vert^2\\ s.t.~&\Vert \beta\Vert^2-c\leq 0\end{align}
The Lagrangian for Problem 2 reads:
\begin{equation}
\mathcal{L}(\beta,\lambda)=\frac{1}{2}\Vert y-X\beta\Vert^2+\lambda (\Vert \beta\Vert^2-c)
\end{equation}
and you probably already see the resemblance with Problem 1 (identical except for the constant term $-\lambda c$).
Now let us look at the necessary conditions for optimality. For Problem 1, these read:
\begin{equation}
\nabla_\beta f_\alpha(\beta^*(\alpha))=0
\end{equation}
where we voluntarily write $\beta^*(\alpha)$ to show that this is the optimal solution for a given $\alpha$.
For Problem 2, the KKT conditions imply that we have:
\begin{align*}
\nabla_\beta \mathcal{L}(\beta^*,\lambda^*)&=\nabla_\beta f_\lambda(\beta^*)=0\\
\lambda^* (\Vert \beta^*\Vert^2-c)&=0
\end{align*}
The first line says that the gradient of the Lagrangian with respect to $\beta$ should be null and the second is the complementary condition. (We also need $\lambda^* \geq 0$, but this is less important for our discussion). Also observe that the gradient of the Lagrangian is equal to the gradient of $f_\lambda$ (objective function of problem 1 but with $\lambda$ instead of $\alpha$).
Now suppose we solve Problem 1 for a given $\alpha$ and obtain its solution $\beta^*(\alpha)$. Let $c=\Vert \beta^*(\alpha)\Vert^2$, the squared norm of the solution to Problem 1. Then $\lambda^*=\alpha$ and $\beta^*=\beta^*(\alpha)$ satisfy the KKT conditions for Problem 2, showing that both Problems have the same solution. Conversely, if you solved Problem 2, you could set $\alpha=\lambda^*$ to retrieve the same solution by solving Problem 1.
To sum it up, both problems are equivalent when $c=\Vert \beta^*(\alpha)\Vert^2$.
