The equation of a sphere

A Wi-Fi router is placed at $I(2, 3, 1)$ in a hall (unit: metre). The signal is strong enough within $10$ m of the router. Does a phone at $M(8, 10, 3)$ get the signal? What about a speaker hanging from the ceiling at $N(2, 3, 12)$ ?

The set of points exactly $10$ m from $I$ is a sphere. This lesson turns a sphere into an equation, so that every "inside or outside" question becomes a calculation.

Exploring: distances from the router

We find the distances from the router to the two points: $IM=\sqrt{6^2+7^2+2^2}=\sqrt{89}\approx9.43<10,\qquad IN=\sqrt{0^2+0^2+11^2}=11>10.$ So the phone gets the signal, but the speaker is outside the coverage area. A point $P(x, y, z)$ lies exactly on the edge of the coverage area when $IP=10$ , that is $(x-2)^2+(y-3)^2+(z-1)^2=100.$

The sphere with centre

I

and radius

R

Why is it true? $M(x, y, z)$ lies on the sphere exactly when $IM^2=R^2$ , and $IM^2=(x-a)^2+(y-b)^2+(z-c)^2$ . For the second form, complete the squares: the equation becomes $(x-a)^2+(y-b)^2+(z-c)^2=a^2+b^2+c^2-d$ . The right-hand side must be positive to be $R^2$ .

Example Given the centre and radius; given a diameter

a) Write the equation of the sphere with centre $I(1, -2, 3)$ and radius $4$ , and check that $A(1, 2, 3)$ lies on it.

b) Write the equation of the sphere with diameter $AB$ , where $A(1, 3, -1)$ , $B(3, -1, 5)$ .

Solution. a) Substitute the centre and the radius into the formula. Note that $y-(-2)=y+2$ :

$(x-1)^2+(y+2)^2+(z-3)^2=16.$

Substituting $A$ : $0^2+4^2+0^2=16$ , so $A$ lies on the sphere.

b) The centre is the midpoint $I$ of $AB$ and the radius is half of $AB$ :

$I(2, 1, 2),\qquad AB=\sqrt{2^2+(-4)^2+6^2}=\sqrt{56},\qquad R=\frac{\sqrt{56}}{2}=\sqrt{14}.$

So a) $(x-1)^2+(y+2)^2+(z-3)^2=16$ ; b) $(x-2)^2+(y-1)^2+(z-2)^2=14$ .

When the equation is expanded, we read off $a$ , $b$ , $c$ by dividing the coefficients of $x$ , $y$ , $z$ by $-2$ .

Example From the expanded form

Is each equation the equation of a sphere? If so, find its centre and radius.

a) $x^2+y^2+z^2-4x+2y-6z-2=0$ . b) $x^2+y^2+z^2-2x+4y+6=0$ .

Solution. a) Divide the coefficients $-4$ , $2$ , $-6$ by $-2$ : $a=2$ , $b=-1$ , $c=3$ ; and $d=-2$ . Check the condition:

$a^2+b^2+c^2-d=4+1+9+2=16>0,\qquad R=4.$

b) $a=1$ , $b=-2$ , $c=0$ , $d=6$ :

$a^2+b^2+c^2-d=1+4+0-6=-1<0.$

So a) is a sphere with centre $I(2, -1, 3)$ and radius $4$ ; b) is not the equation of a sphere (no point satisfies it).

Given four points on a sphere, the expanded form gives four linear equations in the four unknowns $a$ , $b$ , $c$ , $d$ .

Example The sphere through four points

Write the equation of the sphere through $O(0, 0, 0)$ , $A(2, 0, 0)$ , $B(0, 4, 0)$ , $C(0, 0, 6)$ .

Solution. Let the sphere be $x^2+y^2+z^2-2ax-2by-2cz+d=0$ . Substitute the four points in turn:

$\begin{aligned} O&: d=0;\\ A&: 4-4a+d=0\Rightarrow a=1;\\ B&: 16-8b+d=0\Rightarrow b=2;\\ C&: 36-12c+d=0\Rightarrow c=3. \end{aligned}$

Then $R^2=1+4+9-0=14$ . So the sphere is $x^2+y^2+z^2-2x-4y-6z=0$ , with centre $I(1, 2, 3)$ and radius $\sqrt{14}$ .

Planes and spheres

We compare the distance $h$ from the centre $I$ to the plane with the radius $R$ , just as we compare the distance from the centre to a line in plane geometry.

Example Three planes and a sphere

Let $(S): (x-1)^2+(y+1)^2+(z-2)^2=9$ . Decide how $(S)$ meets $(P): 2x-y+2z+2=0$ , $(Q): x+2y+2z-6=0$ , $(R): 2x+2y-z+15=0$ . If a plane touches the sphere, find the point of contact; if it cuts the sphere, find the radius of the circle of intersection.

Solution. The centre is $I(1, -1, 2)$ and $R=3$ . All three normal vectors have length $3$ , so:

$d(I, (P))=\frac{|2+1+4+2|}{3}=3,\qquad d(I, (Q))=\frac{|1-2+4-6|}{3}=1,\qquad d(I, (R))=\frac{|2-2-2+15|}{3}=\frac{13}{3}.$

$(P)$ touches $(S)$ . The point of contact $H$ is the projection of $I$ : the line $x=1+2t$ , $y=-1-t$ , $z=2+2t$ meets $(P)$ when $9+9t=0$ , that is $t=-1$ , so $H(-1, 0, 0)$ .

$(Q)$ cuts $(S)$ in a circle of radius $r=\sqrt{9-1}=2\sqrt2$ . Since $\dfrac{13}{3}>3$ , $(R)$ does not meet $(S)$ .

So $(P)$ touches $(S)$ at $H(-1, 0, 0)$ , $(Q)$ cuts $(S)$ in a circle of radius $2\sqrt2$ , and $(R)$ has no common point with $(S)$ .

The last example is a real problem with several steps, so we follow the four stages.

A vertical cross-section of the radar zone (unit:

10

km)

Example The radar zone

A radar station at $O(0, 0, 0)$ detects aircraft within $50$ km. An aeroplane flies in a straight line at a height of $30$ km along the line $x=-60+t$ , $y=20$ , $z=30$ (unit: km). How many kilometres does the aeroplane fly inside the zone the radar can detect (to 1 decimal place)?

Solution.

Understand. The detection zone is the solid sphere $x^2+y^2+z^2\le2500$ . We need the length of the part of the flight path inside it.

Plan. The flight path lies in the plane $z=30$ . This plane is at distance $h=30<50$ from $O$ , so it cuts the sphere in a circle of radius $r=\sqrt{50^2-30^2}=40$ with centre $(0, 0, 30)$ .

Solve. At height $30$ , the aeroplane is inside the radar zone when $x^2+y^2\le1600$ . With $y=20$ :

$x^2\le1600-400=1200\iff -20\sqrt3\le x\le20\sqrt3.$

This stretch has length $40\sqrt3\approx69.3$ km (each unit of $t$ is $1$ km along the $x$ -axis).

So the aeroplane flies about $69.3$ km inside the radar zone.

Look back. We turned a problem in space into a circle in the plane $z=30$ , thanks to the formula $r=\sqrt{R^2-h^2}$ .

Watch out Common mistakes

The wrong sign for the centre. Wrong: $(x+1)^2+(y-2)^2+z^2=9$ has centre $(1, -2, 0)$ . Why it is wrong: $x+1=x-(-1)$ . Right: the centre is $I(-1, 2, 0)$ .
Confusing $R$ with $R^2$ . Wrong: the sphere with centre $I(1, -2, 3)$ and radius $4$ is $(x-1)^2+(y+2)^2+(z-3)^2=4$ . Right: the right-hand side is $R^2=16$ .
Reading off the centre when the coefficient of $x^2$ is not $1$ . For $2x^2+2y^2+2z^2-4x+8y-2=0$ , wrong: $a=2$ , $b=-4$ . Why it is wrong: the formula needs the coefficients of $x^2$ , $y^2$ , $z^2$ to be $1$ . Right: divide by $2$ to get $x^2+y^2+z^2-2x+4y-1=0$ , so $a=1$ , $b=-2$ , $c=0$ and $R=\sqrt{1+4+0+1}=\sqrt6$ .
Forgetting to check $a^2+b^2+c^2-d>0$ . Wrong: $x^2+y^2+z^2-2x+4y+6=0$ has $R=\sqrt{-1}$ , "undefined but still a sphere". Right: the condition fails, so the equation describes no sphere at all.

Key terms

English	Tiếng Việt
sphere	mặt cầu
centre	tâm
radius	bán kính
diameter	đường kính
solid sphere (ball)	khối cầu
tangent plane	mặt phẳng tiếp diện
point of contact	tiếp điểm
circle of intersection	đường tròn giao tuyến

Summary

Centre $I(a, b, c)$ , radius $R$ : $(x-a)^2+(y-b)^2+(z-c)^2=R^2$ .
$x^2+y^2+z^2-2ax-2by-2cz+d=0$ is a sphere $\iff a^2+b^2+c^2-d>0$ ; then $R=\sqrt{a^2+b^2+c^2-d}$ .
Diameter $AB$ : the centre is the midpoint of $AB$ and $R=\dfrac{AB}{2}$ .
A point $M$ : $IM<R$ inside, $IM=R$ on, $IM>R$ outside the sphere.
A plane: $h=d(I, (P))$ ; $h>R$ no common point, $h=R$ tangent, $h<R$ a circle of radius $\sqrt{R^2-h^2}$ .
The tangent plane at $M$ : through $M$ with normal vector $\overrightarrow{IM}$ .

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Exercise 1

Write the equation of the sphere with centre $I(-1, 2, 0)$ and radius $3$ . Is the point $A(1, 3, 2)$ inside, on or outside the sphere?

What you need and how to start

What you need: $(x-a)^2+(y-b)^2+(z-c)^2=R^2$ ; compare $IA$ with $R$ .

Steps:

Substitute $a=-1$ , $b=2$ , $c=0$ , $R=3$ .
Compute $IA^2$ .

Show solution

$(x+1)^2+(y-2)^2+z^2=9$ . $IA^2=2^2+1^2+2^2=9=R^2$ .

Answer. $(x+1)^2+(y-2)^2+z^2=9$ ; $A$ lies on the sphere.

Exercise 2

Find the centre and radius of the sphere $x^2+y^2+z^2+4x-6y+2z-11=0$ .

What you need and how to start

What you need: $a$ , $b$ , $c$ are the coefficients of $x$ , $y$ , $z$ divided by $-2$ ; $R=\sqrt{a^2+b^2+c^2-d}$ .

Steps:

Read off $a$ , $b$ , $c$ , $d$ .
Compute $R$ .

Show solution

$a=-2$ , $b=3$ , $c=-1$ , $d=-11$ . $R^2=4+9+1+11=25$ .

Answer. $I(-2, 3, -1)$ , $R=5$ .

Exercise 3

Which of these are equations of spheres? Find the centre and radius of each sphere.

a) $x^2+y^2+z^2-2x+4z+1=0$ . b) $x^2+y^2+z^2+2y+5=0$ . c) $2x^2+2y^2+2z^2-4x+8y-2=0$ .

What you need and how to start

What you need: the condition $a^2+b^2+c^2-d>0$ , with the coefficients of $x^2$ , $y^2$ , $z^2$ equal to $1$ .

Steps:

In part c), divide both sides by $2$ first.
Check the condition for each equation.

Show solution

a) $a=1$ , $b=0$ , $c=-2$ , $d=1$ : $1+0+4-1=4>0$ , centre $(1, 0, -2)$ , $R=2$ . b) $a=0$ , $b=-1$ , $c=0$ , $d=5$ : $1-5=-4<0$ , not a sphere. c) $x^2+y^2+z^2-2x+4y-1=0$ : $1+4+0+1=6>0$ , centre $(1, -2, 0)$ , $R=\sqrt6$ .

Answer. a) and c) are spheres; b) is not.

Exercise 4

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

What you need and how to start

What you need: the centre is the midpoint of $AB$ and $R=\dfrac{AB}{2}$ .

Steps:

Find the midpoint $I$ .
Find $AB$ , then $R$ .

Show solution

$I(1, 1, 1)$ , $AB=\sqrt{4+16+16}=6$ , $R=3$ .

Answer. $(x-1)^2+(y-1)^2+(z-1)^2=9$ .

Exercise 5

(Revisiting The equation of a plane.) Write the equation of the sphere with centre $I(1, 2, -1)$ that touches the plane $(P): 2x-2y+z+9=0$ .

What you need and how to start

What you need: the sphere touches $(P)$ when $R=d(I, (P))$ .

Steps:

Find $d(I, (P))$ with the distance formula.
Write the equation with this $R$ .

Show solution

$R=d(I, (P))=\frac{|2-4-1+9|}{\sqrt{4+4+1}}=\frac63=2.$

Answer. $(x-1)^2+(y-2)^2+(z+1)^2=4$ .

Exercise 6

Let $(S): (x-2)^2+(y+1)^2+z^2=9$ . Is each claim true or false?

a) $(S)$ has centre $I(2, -1, 0)$ .

b) $(S)$ has radius $9$ .

c) The point $M(4, 1, 1)$ lies on $(S)$ .

d) The plane $z=4$ cuts $(S)$ in a circle.

What you need and how to start

What you need: read off the centre and radius; compare the distance from the centre to the plane with $R$ .

Steps:

Read off $I$ and $R$ .
Substitute $M$ into the left-hand side.
Find the distance from $I$ to the plane $z=4$ .

Show solution

a) True. $x-2$ , $y+1=y-(-1)$ , $z-0$ .

b) False. $R^2=9$ , so $R=3$ .

c) True. $2^2+2^2+1^2=9$ .

d) False. The distance from $I$ to the plane $z=4$ is $|0-4|=4>3$ , so they have no common point.

Answer. a) True; b) False; c) True; d) False.

Exercise 7

A Bluetooth speaker hangs at $I(2, 3, 4)$ in a room (unit: metre), and the floor is the plane $z=0$ . The speaker can be heard clearly within $6$ m of it. Find the area of the part of the floor where the speaker is heard clearly (in m², to 1 decimal place).

What you need and how to start

What you need: a plane at distance $h<R$ from the centre cuts the sphere in a circle of radius $r=\sqrt{R^2-h^2}$ .

Steps:

$h=4$ , $R=6$ ; find $r$ .
The area of a disc is $\pi r^2$ .

Show solution

$r^2=36-16=20$ , so the area is $20\pi\approx62.8$ m².

Answer. $62.8$ .

Exercise 8

Write the equation of the sphere whose centre lies on the axis $Ox$ and which passes through $A(1, 2, 3)$ and $B(3, 0, 1)$ .

What you need and how to start

What you need: a centre on $Ox$ has the form $I(t, 0, 0)$ ; the condition $IA=IB$ .

Steps:

Write $IA^2=IB^2$ in terms of $t$ .
Solve for $t$ , then find $R^2=IA^2$ .

Show solution

$(1-t)^2+4+9=(3-t)^2+0+1\iff 14-2t=10-6t\iff t=-1.$ $I(-1, 0, 0)$ , $R^2=4+4+9=17$ .

Answer. $(x+1)^2+y^2+z^2=17$ .

Exercise 9

(Revisiting The equation of a line in space.) Find where the line $d: x=1+t$ , $y=2$ , $z=2-3t$ meets the sphere $(S): x^2+y^2+z^2=9$ , and find the length of the chord.

What you need and how to start

What you need: substituting the parametric equations into the equation of the sphere gives a quadratic equation in $t$ .

Steps:

Substitute $x$ , $y$ , $z$ into $(S)$ and simplify.
Solve for $t$ , find the two points, then the distance between them.

Show solution

$(1+t)^2+4+(2-3t)^2=9\iff 10t^2-10t=0\iff t=0\ \text{or}\ t=1.$ The two points are $(1, 2, 2)$ and $(2, 2, -1)$ . The chord has length $\sqrt{1^2+0^2+3^2}=\sqrt{10}$ .

Answer. $(1, 2, 2)$ , $(2, 2, -1)$ ; the chord is $\sqrt{10}\approx3.16$ long.

Exercise 10 Challenge

Model the Earth as a sphere with centre $O$ and radius $6400$ km. With one thousand km as the unit: $(S): x^2+y^2+z^2=40.96$ . A satellite is at $V(0, 0, 8)$ . The points $M$ on the ground that see the satellite on the horizon are those for which $VM$ is tangent to the sphere, that is $\overrightarrow{OM}\perp\overrightarrow{MV}$ . Prove that these points lie in one plane and find the radius of the circle bounding the coverage area (in km).

What you need and how to start

What you need: $\overrightarrow{OM}\cdot\overrightarrow{MV}=0$ ; $OM^2=40.96$ ; $r=\sqrt{R^2-h^2}$ .

Steps:

Write $\overrightarrow{MV}=\overrightarrow{OV}-\overrightarrow{OM}$ and expand $\overrightarrow{OM}\cdot\overrightarrow{MV}=0$ .
Deduce $8z=40.96$ , then find $r$ .

Show solution

$\overrightarrow{OM}\cdot(\overrightarrow{OV}-\overrightarrow{OM})=0\iff\overrightarrow{OM}\cdot\overrightarrow{OV}=OM^2\iff 8z=40.96\iff z=5.12.$ So these points lie in the plane $z=5.12$ , at distance $h=5.12$ from $O$ : $r=\sqrt{40.96-5.12^2}=\sqrt{14.7456}=3.84.$

Answer. The plane $z=5.12$ ; the boundary circle has radius $3840$ km.

The equation of a sphere