Distance Formula in $3$-space: the distance between points $P(x_1,\,y_1,\, z_1)$ and $Q(x_2,\,y_2,\, z_2)$ is given by $$ \hbox{dist}(P,\,Q) \ = \ \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}\,.$$

With this distance formula, spheres and cylinders can be defined in $3$-space just as circles were defined using the distance formula in the plane:

I. SPHERE: A sphere consists of all points that are a fixed distance $R$, called the radius from a point, called the center of the sphere. When the center is the point $C(a,\,b,\,c)$, then the sphere of $R$ centered at $C$ consists of all points $P(x,\,y,\,z)$ such that $$(x-a)^2 + (y-b)^2 + (z-c)^2 \ = \ R^2\,.$$ The sphere to the right is centered at the origin. The blue circle $x^2+y^2 = R^2$ in the $xy$-plane has been included also. Notice that the blue sphere intersects the $xy$-plane in this circle $x^2+y^2=R^2$.     II. CYLINDER: A (circular) cylinder consists of all points that are a fixed distance from a line, called the axis of symmetry of the cylinder; circular tells us that all its cross-sections perpendicular to this axis of symmetry are circles. When the circle is replaced by an ellipse we get what's called an elliptic cylinder (not surprisingly). The cylinder to the right has the $z$-axis as axis of symmetry, and if the cylinder intersects the $xy$-plane in the green circle $x^2 + y^2 = R^2$, then the cylinder consists of all points $P(x,\,y,\,z)$ such that $$x^2 + y^2 \ = \ R^2\,.$$ Can you see how this equation changes if the $x$-axis or the $y$-axis is the axis of symmetry?

Now use the sliders to change the position of the sphere or the cylinder. When the center of the sphere lies at the origin on the axis of symmetry of the cylinder, notice that the cylinder always touches the sphere at the circle of intersection with the $xy$-plane. We say the cylinder circumscribes the sphere.

Example 2: Find the trace on the $yz$-plane of the sphere having center $(-3,\,3,\,1)$ and radius $4$; in other words, find the intersection on the $yz$-plane of the sphere having center $(-3,\,3,\,1)$ and radius $4$. Solution: The sphere consists of all points $P(x,\,y,\,z)$ such that $$\hbox{dist}\big\{\,(x,\,y,\,z), \, (-3,\,3,\,1)\big\} \ = \ 4\,.$$ So by the distance formula in $3$-space, the coordinates $(x,\,y,\,z)$ satisfy the equation

$$(x+3)^2+(y-3)^2 + (z-1)^2 \ = \ 16\,,$$ which after expansion becomes $$x^2+y^2+z^2 +6x -6y -2z + 3\ = \ 0\,.$$ But the $yz$-plane is the plane $x = 0$. Thus the trace of the sphere on the $yz$-plane is obtained by setting $x = 0$. This is the circle $$y^2 +z^2 -6y -2z + 3\ = \ 0 \,,$$ or equivalently $$(y-3)^2 + (z-1)^2 \ = \ 7\, .$$