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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

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

(<=): Given that $D^2

I believe there is something important that I'm not seeing. Would definitely appreciate a hint.

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    Hint: consider the distance from the origin to the plane. – 2017-01-25
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    I think it's just $z$. – 2017-01-25
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    The distance can not depend on $z$. It is in fact $\left|D / \sqrt{A^2+B^2+C^2}\right|\,$, see for example [here](https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_plane#Restatement_using_linear_algebra). Now, for example, if that distance is greater than $1$, then the plane doesn't intersect the unit sphere at all. Then there are two other cases to consider. – 2017-01-25
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    The point you are missing is that if you rotate the sphere (and the axes with it) then the plane will no longer have the same equation. So you will not (at least not easily) get any condition involving $A,B,C,D$. – 2017-01-25
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    @dxiv such a simple piece makes all the difference. Thanks :) – 2017-01-25

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What dxiv says is the same yours, the distance origin from plane must be less than $1$ or $$\frac{|D|}{\sqrt{A^2+B^2+C^2}}<1$$

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    In your formula, $D$ should be $|D|$ which give a condition equivalent, in fact, to the condition $D^2 – 2017-01-25