Let $B$ be a ball of radius $r$ whose diameter is defined by a line segment between the points $x,y$. Let $B_1, B_2$ be the two balls in the pencil of circles defined by the line segment $xy$ with radii $1/\kappa$. Let $\gamma \subset B$ be a smooth curve contained within the ball $B$ and with endpoints $x$ and $y$. Suppose that $r < \frac{1}{\kappa}$. I wish to show that if the curvature of $\gamma$ is bounded by $\kappa$ everywhere, then $\gamma$ must lie in $B_1 \cap B_2$ the intersection between $B_1, B_2$. See Figure below.
I've made several attempts to prove this statement. Intuitively it seems like if $\gamma$ leaves the $B_1 \cap B_2$ it has to curve more rapidly than $\kappa$ would allow to make it back to $y$. The arc of $B_{2}$ from $x$ to $y$ has curvature everywhere at most $\kappa$. I've tried using osculating circles, rates of change of the derivative of $\gamma$ relative to the arc, assuming a bound on the distance of $\gamma$, assuming a bound on $r$, but I can't seem to make anything work. Any help would be appreciated.
