Conditions for intersection of a plane and a sphere

Question

I want to find a condition in order that the plane $ax+by+cz=d$ and sphere $x^2+y^2+(z-1)^2=1$ has intersection.

Somewhere I read without proof that the plane $ax+by+cz=d$ has intersection with the sphere $x^2+y^2+z^2=1$ if $a^2+b^2+c^2>d^2$.

So I've tried to substitude $z=\frac{d-ax-by}c$ into the equation of sphere. I arrived to equation of a circle

$$(a^2+c^2)x^2+(b^2+c^2)y^2+(2ac-2ad)x+...=0$$

But I don't know how to obtain suitable condition.

Could someone help me or refer me to some book that completely has discussed conditions for intersection of plane and sphere? Thanks!

Answer 1

The centre of the sphere is at $(0,0,1)$ and radius of it is $1$. Plus, if you apply the formula $$\text{dist}=\frac{|c-d|}{\sqrt{a^2+b^2+c^2}}\leq1$$ The condition that the plane $ax+by+cz=d$ has intersection with the sphere $x^2+y^2+z^2=1$ if $a^2+b^2+c^2>d^2$ could be also derived from the distance formula.

By substituting $z=\frac{d-ax-by}c$ into the equation of sphere, you'll get a range of equations of circle. It doesn't generate the condition "when do they intersect". To find such condition/relationship between coefficients, you'll have to find some concrete constraints, subsequently expressing them in equations and so on.

Answer 2

HINT

Note that to have intersection we need that the radius of the circle we have found has radius $R\ge 0$.

As an alternative let consider a translation of the system such that the sphere is centered at the origin and then set the distance of the plane form the origin $\le 1$.

Answer 3

The distance of a point $(p,q,s)$ from a plane $ax+by+cz+d=0$ is $$\frac{|ap+bq+cs+d|}{\sqrt{a^2+b^2+c^2}}$$ (If you omit the absolute value, you will get positive distances from one side of the plane and negative from the other one, which may be useful in some applications, but not in this one.)

So if you apply the formula to the sphere's center, you'll be able to check if the distance of the center from the plane exceeds the sphere's radius or not, hence you will know if they intersect.