= Solution
Set $\mathbf b_0=\mathbf c+h\mathbf n$, where $h=p-\mathbf n\cdot\mathbf c$. Then $\mathbf n\cdot\mathbf b_0=p$, so $\mathbf b_0$ lies in the <plane>. For any position $\mathbf r$ in that <plane>, $\mathbf r-\mathbf b_0$ is perpendicular to $\mathbf b_0-\mathbf c=h\mathbf n$. The <Pythagorean theorem> therefore gives
$$
|\mathbf r-\mathbf c|^2=|\mathbf r-\mathbf b_0|^2+h^2.
$$
On the <sphere> the left side is $a^2$. Thus the intersection is the <circle> in the <plane> centred at $\mathbf b_0$, with squared radius $a^2-h^2$. Its centre must therefore be $\mathbf b=\mathbf b_0$, and
$$
\boxed{\mathbf b-\mathbf c=\lambda\mathbf n,\qquad
\lambda^2=a^2-\rho^2.}
$$
In terms of the radii, \b[$\lambda=\pm\sqrt{a^2-\rho^2}$], with the sign fixed by $\lambda=p-\mathbf n\cdot\mathbf c$. In the central-plane case $\rho=a$, both signs give zero. This is the <center and radius of a sphere-plane intersection> formula with its orientation made explicit.
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