This question is derived from a problem about conjugate points in Riemannian geometry. I have some doubt on the solution, so I'm proposing the most confusing part of it as a question.
Let $M$ be a closed subset of $\Bbb{R}^n$ consisting of disjoint embedded smooth boundless manifolds, each with codimension at least $1$. (In the original problem, $M$ is the pre-image of one point in the conjugate pair under the exponential map w.r.t. the other point) Suppose the following $2$ properties are satisfied:
$1$. Any ray starting from $O$ intersects $M$;
$2$. If the ray $\overrightarrow{OP}$ intersects $M$ at $P$, then there exists a circle $C$, centered at $O$ and passing through $P$, contained in $M$ (not necessarily a whole connected component).
Must $M$ contain a sphere $S$ centered at $O$?
Consider $\mathbb R^4 = \mathbb C^2$ and the equation $$ \|z_1\|^2 + 2\|z_2\|^2 = 1. $$ The set $M = \{\|z_1\|^2 + 2\|z_2\|^2 = 1\}$ is a manifold by the regular value theorem. If $(z_1, z_2) \in \mathbb C^2 \setminus \{0\}$ then for $\lambda \in \mathbb R$, $$ f(\lambda) = \|\lambda z_1\|^2 + 2 \|\lambda z_2\|^2 = \lambda^2f(1) $$ is a continuous function which goes to $0$ as $\lambda \to 0$ and $\infty$ as $\lambda \to \infty$ and hence there is some $\lambda$ for which $f(\lambda) = 1$. Hence any ray from the origin intersects $M$. For the circles note that $e^{i\theta}(z_1, z_2) \in M$ if and only if $(z_1, z_2) \in M$. Note also that any ray intersects $M$ at only one point. Hence $M$ does not contain a sphere, if it did then all the points of $M$ would be at the same distance from the origin.