Chapter 9.3, Problem 19E

(a)

To determine

To find:The trace of the sphere in the $yz-$ plane.

Answer to Problem 19E

The trace of the sphere in the $yz-$ plane is circle ${\left(y-2\right)}^2+{\left(z+10\right)}^2={\left(\sqrt{99}\right)}^2,x=0$ .

Explanation of Solution

Given information:

The given equation of the sphere is ${\left(x+1\right)}^2+{\left(y-2\right)}^2+{\left(z+10\right)}^2=100$ .

Calculation:

As we know that in $yz-$ plane the $x$ coordinate is zero. So, the trace of the sphere in $yz-$ plane consist of all points on the sphere whose $x$ coordinate is zero.

Substitute $0$ for $x$ in given equation.

$\begin{array}{c}{\left(0+1\right)}^2+{\left(y-2\right)}^2+{\left(z+10\right)}^2=100\\ 1+{\left(y-2\right)}^2+{\left(z+10\right)}^2=100\\ {\left(y-2\right)}^2+{\left(z+10\right)}^2=99\end{array}$

The trace of the sphere is circle ${\left(y-2\right)}^2+{\left(z+10\right)}^2={\left(\sqrt{99}\right)}^2,x=0$ .

The radius of the circle is $\sqrt{99}$ , which is in $yz-$ plane, centered at $\left(0,2,-10\right)$ .

Therefore, thetrace of the sphere in the $yz-$ planeis circle ${\left(y-2\right)}^2+{\left(z+10\right)}^2={\left(\sqrt{99}\right)}^2,x=0$ .

(b)

To determine

To find:The trace of the sphere in the plane $x=4$ .

Answer to Problem 19E

The trace of the sphere in the plane $x=4$ is circle ${\left(y-2\right)}^2+{\left(z+10\right)}^2={\left(\sqrt{75}\right)}^2,x=4$ .

Explanation of Solution

Given information:

The given equation of the sphere is ${\left(x+1\right)}^2+{\left(y-2\right)}^2+{\left(z+10\right)}^2=100$ .

Calculation:

As we know that in plane consist of all points on the sphere whose $x$ coordinate is $4$ .

Substitute $4$ for $x$ in given equation.

$\begin{array}{c}{\left(4+1\right)}^2+{\left(y-2\right)}^2+{\left(z+10\right)}^2=100\\ {\left(5\right)}^2+{\left(y-2\right)}^2+{\left(z+10\right)}^2=100\\ 25+{\left(y-2\right)}^2+{\left(z+10\right)}^2=100\\ {\left(y-2\right)}^2+{\left(z+10\right)}^2=75\end{array}$

The trace of the sphere is circle ${\left(y-2\right)}^2+{\left(z+10\right)}^2={\left(\sqrt{75}\right)}^2,x=4$ .

The radius of the circle is $\sqrt{75}$ , that is four units to the right of $yz-$ plane, centered at $\left(4,2,-10\right)$ .

Therefore, the trace of the sphere in the plane $x=4$ is circle ${\left(y-2\right)}^2+{\left(z+10\right)}^2={\left(\sqrt{75}\right)}^2,x=4$ .