Radius of the section

Statement

For the sphere $x^2+y^2+z^2-4x+2y-6z+5=0$ (centre $C(2;-1;3)$, radius $R=3$), calculate the radius of the section with the plane $z=0$.

Solution

The distance of the centre from the plane $z=0$ is $d=|z_C|=3$. Since $d=R=3$, the plane is tangent to the sphere: the “section” degenerates into a single point and the radius of the circle is $r=\sqrt{R^2-d^2}=\sqrt{9-9}=\boxed{0}.$ The plane $z=0$ touches the sphere at the single point $(2;-1;0)$.

Links

Topics: Analytic geometry in space
Concepts: Point-plane distance · Cartesian plane · Circular section · Sphere
Skills: Analytic geometry · Using formulae
Exercise types: Geometric problem