Sphere and plane section and tangent plane

Text

Given the sphere $S:\ x^2+y^2+z^2-2x+4y-2z-10=0$ and the plane $\pi:\ 2x-y+2z-3=0$.

(a) Find the centre $C$ and radius $r$.

(b) Compute $d(C,\pi)$ and determine whether $\pi$ is secant, tangent, or external with respect to $S$.

(c) Find the radius and centre of the section circle.

(d) Verify that $P(5,-2,1)\in S$ and write the plane tangent to $S$ at $P$.

Solution

(a) Completing the squares: $(x-1)^2+(y+2)^2+(z-1)^2=16\Rightarrow C(1,-2,1),\ r=4$.

(b) $d(C,\pi)=\dfrac{|2+2+2-3|}{3}=1$; since $d<r$, the plane is secant.

(c) The radius of the section is $\sqrt{16-1}=\sqrt{15}\approx3.873$; the centre of the section is $C'=C-d\cdot\dfrac{\vec n}{|\vec n|}=\left(\tfrac13,-\tfrac53,\tfrac13\right)$.

(d) For $P(5,-2,1)$: $(5-1)^2+(-2+2)^2+(1-1)^2=16$ ✓, so $P\in S$. The normal vector to the tangent plane at $P$ is $\overrightarrow{CP}=(4,0,0)$, so the tangent plane is $x=5$. $\boxed{C(1,-2,1),\ r=4;\ d=1\ (\text{secant});\ \rho_{\text{sez}}=\sqrt{15}\approx3.87,\ C'\left(\tfrac13,-\tfrac53,\tfrac13\right);\ \text{tangent at } P:\ x=5}$