Consider $\mathbb S_R^2 \subseteq \mathbb R^3$ be a sphere of radius $R$. Let $N$ denote the north pole and $0 < r < R \pi$. Define the metric circle with center N to be all points, $x \in \mathbb S_R^2$ that $$d_{\mathbb S_R^2}(x,N) = r$$ Here $d_{\mathbb S_R^2}$ is the induced metric on the sphere. Parameterize $\mathbb S_R^2$ with the usual spherical coordinates:
$$\Gamma:[0,2\pi) \times [0,\pi]\to\mathbb{R}^3 \quad \quad (\phi, \theta)\mapsto (R\sin\theta\cos\phi,R\sin\theta\sin\phi, R\cos\theta),$$
My hunch is that the metric circle with center N is spherical circle: $$\Gamma ( [0,2\pi) \times \{ r / R \} ) \subseteq \mathbb S_R^2$$ However, I cannot justify this claim using only elementary arguments using only elementary techniques.
Edit: Bump!
Consider sphere $S(0,R)$ of radius $R$ in the Euclidean 3-space $E^3$ and centered at the origin $0$. There are two metrics one defines on $S(0,R)$:
The chordal metric $d(P,Q)=a=|P-Q|$, the restriction of the Euclidean distance function to the sphere.
The angular metric $\alpha=\angle(P,Q)$ which is the angle of the triangle $P0Q$ at $0$ (if you like, you can multiply it by the radius $R$).
Let us express one metric in terms of the other. For this we use the cosine formula: $$ a^2= 2R^2 - 2R^2 \cos(\alpha)= 2R^2(1- \cos(\alpha)). $$ Since $0\le \alpha\le \pi$, it follows that $\alpha$ determines $a$ and vice versa: $$ \alpha= \arccos(1-\frac{a^2}{2R^2}) $$ Now, consider the circle $C(P,\alpha)\subset S(0,R)$ of radius $\alpha$ centered at $P$, where the radius of the circle is computed in terms of the angular metric. Then a point $Q\in S(0,R)$ lies on $C(P,\alpha)$ if and only if $$ a:=d(P,Q)= R\sqrt{2(1-\cos(\alpha))}. $$ In other words, $C(P,\alpha)$ is the same as the circle in $S(0,R)$ of the chordal radius $a$ and centered at $P$, answering your question. (Your question has a typo, I think, you meant $r=p$, but, regardless, your conjectural formula for the angular radius in terms of the chordal radius is wrong.)