$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Lets $\ds{\vec{p}}$ the position of point $P$. For a given $\ds{\hat{n}}$, it intersects the circle if
$\ds{\verts{\vec{p} + \mu\hat{n}}^{2} = a^{2}}$ for some value of $\ds{\mu \in \mathbb{R}}$:
\begin{align}
\mu^{2} + 2\vec{p}\cdot\hat{n}\,\mu + p^{2} = a^{2} & \imp\
\left\lbrace\begin{array}{rcl}
\ds{\mu} & \ds{=} &
\ds{-\vec{p}\cdot\hat{n}}
\\[1mm]
\ds{\pars{\vec{p}\cdot\hat{n}}^{2} + a^{2}} & \ds{=} & \ds{p^{2}}
\end{array}\right.\tag{1}
\end{align}
There are $\ds{\ul{two}}$ solutions $\ds{\hat{n}_{\pm}}$ for $\ds{\hat{n}}$. So, the intersection ocurrs at two points:
$\ds{\vec{p} - \vec{p}\cdot\hat{n}_{-}\,\hat{n}_{-}}$ and
$\ds{\vec{p} - \vec{p}\cdot\hat{n}_{+}\,\hat{n}_{+}}$. Then, the locus definition yields
$$
0 =
\bracks{\vec{p} - \pars{\vec{p} - \vec{p}\cdot\hat{n}_{-}\,\hat{n}_{-}}}\cdot
\bracks{\vec{p} - \pars{\vec{p} - \vec{p}\cdot\hat{n}_{+}\,\hat{n}_{+}}} =
\pars{\vec{p}\cdot\hat{n}_{-}}\pars{\vec{p}\cdot\hat{n}_{+}}
\hat{n}_{-}\cdot\hat{n}_{+}
$$
However, as we can see from expressions $\pars{1}$,
$\ds{\vec{p}\cdot\hat{n}_{\pm} \not= 0}$ because $\ds{a^{2} \not= p^{2}}$ which leads to $\ds{\hat{n}_{-}\cdot\hat{n}_{+} = 0}$
Therefore, $\ds{\vec{p}}$ belong to the OP mentioned locus if there exists a pair of unit vectors $\ds{\hat{n}_{\pm}}$ such that:
\begin{equation}
\left\lbrace\begin{array}{rcl}
\ds{\pars{\vec{p}\cdot\hat{n}_{-}}^{2} + a^{2}} & \ds{=} & \ds{p^{2}}
\\
\ds{\pars{\vec{p}\cdot\hat{n}_{+}}^{2} + a^{2}} & \ds{=} & \ds{p^{2}}
\\
\ds{\hat{n}_{-}\cdot\hat{n}_{+}} & \ds{=} & \ds{0}
\end{array}\right.\tag{2}
\end{equation}
In writing $\ds{\vec{p}}$ as a linear combination of $\ds{\braces{\hat{n}_{\pm}}}$; namely,
$\ds{\vec{p} = c_{-}\hat{n}_{-} + c_{+}\hat{n}_{+}}$; we
find $\ds{c_{\pm}^{2} = a^{2}}$
$\ds{\pars{~\mbox{see conditions}\ \pars{2}~}}$ such that
$$
\color{#f00}{p} = \verts{\vec{p}} = \color{#f00}{\root{2}a}
$$
The locus is the set of points that rest in the circle
$\color{#f00}{\ds{x^{2} + y^{2} = 2a^{2}}}$.
