The sphere and its position relative to points, lines, planes and spheres – Points, Straight Lines and Shapes in 3D (Senior 5 Mathematics) | Ganzaa

Points, Straight Lines and Shapes in 3D

Lessons

Senior 5 · Mathematics

Points, Straight Lines and Shapes in 3D

5 lessons

Lesson 5 of 5

The sphere and its position relative to points, lines, planes and spheres

40 min

All the points at one distance

A football, a water tank on a hill, a planet: each is (close to) a sphere, the set of points at a fixed distance from a centre. GPS receivers use spheres too: knowing its distance from one satellite places a phone somewhere on a sphere around that satellite, and the spheres around several satellites meet in very few points. Sphere problems in this lesson come down to one idea from the earlier lessons: compare a distance with the radius.

Definition

The sphere with centre $C (a, b, c)$ and radius $r > 0$ is the set of points $P$ with $C P = r$ :

(x - a)^{2} + (y - b)^{2} + (z - c)^{2} = r^{2} .

Expanded, it has the general form $x^{2} + y^{2} + z^{2} + 2 u x + 2 v y + 2 w z + d = 0$ , with centre $(- u, - v, - w)$ and radius $\sqrt{u^2 + v^2 + w^2 - d}$ . Such an equation is a sphere only if $u^{2} + v^{2} + w^{2} - d > 0$ : the coefficients of $x^{2}, y^{2}, z^{2}$ must be equal and there is no $x y$ , $y z$ or $z x$ term.

Worked example 1: centre and radius by completing squares

Find the centre and radius of $x^{2} + y^{2} + z^{2} - 4 x + 6 y - 2 z - 2 = 0$ .

Solution. $(x - 2)^{2} - 4 + (y + 3)^{2} - 9 + (z - 1)^{2} - 1 - 2 = 0$ , so $(x - 2)^{2} + (y + 3)^{2} + (z - 1)^{2} = 16$ . Centre $(2, - 3, 1)$ , radius 4.

Worked example 2: a sphere on a diameter

Find the equation of the sphere with diameter $A B$ , where $A (1, 2, - 1)$ and $B (5, 0, 3)$ .

Solution. The centre is the midpoint $(3, 1, 1)$ and the radius is $\frac{AB}{2} = \frac{\sqrt{16 + 4 + 16}}{2} = 3$ :

(x - 3)^{2} + (y - 1)^{2} + (z - 1)^{2} = 9.

Important

Positions relative to a sphere $S$ (centre $C$ , radius $r$ ). In every case, compare a distance with $r$ .

Object	Distance to compare	Less than $r$	Equal to $r$	Greater than $r$
point $P$	$C P$	inside	on $S$	outside
line $L$	$d (C, L)$	meets $S$ in 2 points	tangent	no common point
plane $Π \Pi$	$\delta = d(C, \Pi)$	cuts $S$ in a circle	tangent plane	no common point

For a line you may instead substitute its parametric equations into the sphere's equation: a quadratic in $t$ with two, one or no real roots.

When a plane cuts the sphere, the circle of intersection has its centre at $H$ , the foot of the perpendicular from $C$ to $Π \Pi$ , and radius $\rho = \sqrt{r^2 - \delta^2}$ , by Pythagoras in the right triangle $C H Q$ of the figure. The tangent plane at a point $T$ of $S$ is the plane through $T$ with normal $\overrightarrow{CT}$ .

Two spheres with centres at distance $D$ and radii $r_{1}, r_{2}$ : they meet in a circle if $∣ r_{1} - r_{2} ∣ < D < r_{1} + r_{2}$ ; touch if $D = r_{1} + r_{2}$ (externally) or $D = ∣ r_{1} - r_{2} ∣$ (internally); otherwise they have no common point.

Worked example 3: a plane cutting a sphere

Show that the plane $2 x - y + 2 z = 3$ cuts the sphere of worked example 1 in a circle, and find the circle's centre and radius.

Solution. $C (2, - 3, 1)$ , $r = 4$ , $n = (2, - 1, 2)$ , $∣ n ∣ = 3$ . $\delta = \frac{|4 + 3 + 2 - 3|}{3} = 2 < 4$ , so the plane cuts the sphere. The radius is $\rho = \sqrt{16 - 4} = 2\sqrt{3} \approx 3.46$ .

The centre $H$ is on the line $r = C + t\,n$ through $C$ perpendicular to the plane: $2 (2 + 2 t) - (- 3 - t) + 2 (1 + 2 t) = 3$ gives $9 + 9 t = 3$ , $t = -\frac{2}{3}$ , so $H = \left(\frac{2}{3}, -\frac{7}{3}, -\frac{1}{3}\right)$ . Check: $CH = \frac{2}{3}|n| = 2 = \delta$ . ✓

Worked example 4: a line through a sphere, and tangent planes

(a) Find where the line $r = (1, 2, 2) + t (1, - 4, - 1)$ meets the sphere $x^{2} + y^{2} + z^{2} = 9$ . (b) For which $k$ is the plane $x + 2 y + 2 z = k$ tangent to the sphere $(x - 3)^{2} + (y - 1)^{2} + (z - 1)^{2} = 9$ ?

Solution. (a) $(1 + t)^{2} + (2 - 4 t)^{2} + (2 - t)^{2} = 9$ simplifies to $18 t^{2} - 18 t = 0$ , so $t = 0$ or $t = 1$ : the points $(1, 2, 2)$ and $(2, - 2, 1)$ . Both satisfy $1 + 4 + 4 = 4 + 4 + 1 = 9$ . ✓

(b) Tangent means $\delta = r$ : $\frac{|3 + 2 + 2 - k|}{3} = 3$ , so $∣ 7 - k ∣ = 9$ and $k = - 2$ or $k = 16$ . Two parallel tangent planes, one on each side of the sphere.

Worked example 5: two spheres

How do the spheres $S_{1}$ , centre $(2, - 3, 1)$ , radius 4, and $S_{2}$ , centre $(3, 1, 1)$ , radius 3, lie relative to each other?

Solution. $D = \sqrt{1 + 16 + 0} = \sqrt{17} \approx 4.12$ . Since $∣ 4 - 3 ∣ = 1 < 4.12 < 7 = 4 + 3$ , the spheres meet in a circle.

Warning

Halve the coefficients: in $x^{2} - 4 x$ , the centre's coordinate is $+ 2$ , not $- 4$ or $4$ .
Not every "sphere-like" equation is a sphere: $x^{2} + y^{2} + z^{2} - 2 x + 4 y + 6 = 0$ gives $r^{2} = 1 + 4 - 6 = - 1$ , so it has no points at all.
The circle of intersection has radius $\sqrt{r^2 - \delta^2}$ , not $r - \delta$ .

Tip

Every sphere question in this lesson is the same question: "how far is the centre?" Find that distance first, then compare it with the radius.

Mathispower4u: The Equation of a Sphere (about 10 min). Worked examples of finding the centre and radius by completing the square, as in worked example 1.

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Practice

Questions 1 to 4 are routine; 5 to 7 need more care; 8 asks for a reason.

Write the equation of the sphere with centre $(1, - 2, 3)$ and radius 4 in general form.
Find the centre and radius of $x^{2} + y^{2} + z^{2} + 2 x - 4 y + 8 z + 5 = 0$ .
Is $P (1, 1, 1)$ inside, on or outside the sphere of question 2?
Is $x^{2} + y^{2} + z^{2} - 2 x + 4 y + 6 = 0$ a sphere?
Describe the intersection of $x^{2} + y^{2} + z^{2} = 25$ with the plane $z = 3$ .
Find $k$ so that the plane $x + y + z = k$ is tangent to the sphere $(x - 1)^{2} + y^{2} + (z + 2)^{2} = 4$ .
Show that the spheres $x^{2} + y^{2} + z^{2} = 4$ and $(x - 3)^{2} + y^{2} + z^{2} = 1$ touch, and find the point of contact.
Explain why the tangent plane to a sphere at $T$ is perpendicular to the radius $C T$ .

Show answersHide answers

$x^{2} + y^{2} + z^{2} - 2 x + 4 y - 6 z - 2 = 0$ .
Centre $(- 1, 2, - 4)$ , radius $\sqrt{1 + 4 + 16 - 5} = 4$ .
$CP = \sqrt{4 + 1 + 25} = \sqrt{30} \approx 5.48 > 4$ : outside.
No: $u^{2} + v^{2} + w^{2} - d = 1 + 4 - 6 = - 1 < 0$ .
$\delta = 3 < 5$ : a circle with centre $(0, 0, 3)$ and radius $\sqrt{25 - 9} = 4$ .
$\frac{|1 + 0 - 2 - k|}{\sqrt{3}} = 2$ , so $|{-1} - k| = 2\sqrt{3}$ and $k = -1 \pm 2\sqrt{3}$ .
$D = 3 = 2 + 1 = r_{1} + r_{2}$ : they touch externally, at $(2, 0, 0)$ , the point 2 units from $O$ towards $(3, 0, 0)$ .
Every point of the tangent plane other than $T$ is outside the sphere, so $T$ is the point of the plane nearest to $C$ . The nearest point of a plane is the foot of the perpendicular, so $C T$ is perpendicular to the plane.

1All the points at one distance

All the points at one distance