Let $Ax+By+Cz=D$ be a plane and $S^2$ be a unit sphere in $\mathbb{R}^3$. Prove that the intersection of the plane with $S^2$ is a circle if and only if $D^2<A^2+B^2+C^2$.
I understand that we can rotate the sphere about the origin in such a way as to make the plane parallel to the $xy$-plane. Then we need to prove the following:
(=>): Given that the intersection is a circle, we have that $x^2+y^2=1-z^2$, so $z^2$ must be a constant less than $1$. We can derive $z$ from the plane equation: $z=\frac{D-Ax-By}{C}$, where $Ax+By=$ constant. We need to somehow derive the condition from $D-Ax-By<C$. I haven't been successful with this.
(<=): Given that $D^2<A^2+B^2+C^2$, we need show that $z^2<1$. But I wasn't able to do this either.
I believe there is something important that I'm not seeing. Would definitely appreciate a hint.
What dxiv says is the same yours, the distance origin from plane must be less than $1$ or $$\frac{|D|}{\sqrt{A^2+B^2+C^2}}<1$$