The problem is clear if the two planes are parallel, thus we assume that they intersect on line $l$. Let $r$ denote the radius of the sphere and let $O$ denote the center of the sphere. Let $a > r$ denote the distance from $O$ to $l$.

A sphere with center $P$ and radius $s$ satisfies the condition of the problem iff
$$r + s = OP \quad \text{and} \quad \frac{r}{a} = \frac{s}{b}.$$
We may assume that $r = 1$. Consider a coordinate system where $l$ is the $x$-axis and $O(0, a, 0)$. Then $P(x, y, z)$ satisfies $y = b > 0$ and $z = 0$. Since $a > 1$, the equation
$$x^2 + (y - a)^2 = \left(1 + \frac{y}{a}\right)^2$$
gives an ellipse.