Sphere centre radius and tangent plane

Statement

Given the sphere $x^2+y^2+z^2-4x+2y-6z+5=0$, determine its centre and radius, and write the tangent plane at the point $D$ with $x_D=3$, $z_D=1$.

Solution

Centre and radius. With $a=-4,\ b=2,\ c=-6,\ d=5$: $C\left(-\tfrac{a}{2};-\tfrac{b}{2};-\tfrac{c}{2}\right)=(2;-1;3), \qquad R=\sqrt{4+1+9-5}=\sqrt9=3.$

Point $D$. Substituting $x_D=3,\ z_D=1$ into the equation: $9+y^2+1-12+2y-6+5=0\Rightarrow y^2+2y-3=0\Rightarrow (y-1)(y+3)=0$, so $y=1$ or $y=-3$. We take $D(3;1;1)$.

Tangent plane. $\vec{CD}=D-C=(1;2;-2)$; the tangent plane at $D$ is $(\vec S-\vec D)\cdot\vec{CD}=0$: $1(x-3)+2(y-1)-2(z-1)=0 \iff \boxed{x+2y-2z-3=0}.$ (With the other choice $D'(3;-3;1)$ one obtains a second tangent plane analogously.)

Links

Topics: Analytic geometry in space
Concepts: Tangent plane · Sphere
Skills: Analytic geometry · Using formulae
Exercise type: Geometric problem