The sphere as a locus

What you will be able to do

• write the equation of a sphere from its centre and radius, and read a centre and radius from an equation;
• find the sphere that has two given points as the ends of a diameter;
• find where a sphere meets an axis or a coordinate plane.

A balloon on a string

A balloon is tied to a peg by a string $5$ metres long. The string is always taut. Where can the balloon be? In a plane the answer would be a circle. In space it is a sphere. A locus is the set of all points that satisfy one condition. The locus of points at a fixed distance $r$ from a fixed point $C$ is the sphere with centre $C$ and radius $r$.

The equation of a sphere

Let the centre be $C(a, b, c)$. A point $P(x, y, z)$ is on the sphere when $CP = r$. By the distance formula of the last course, $CP^2 = r^2$ gives

With the centre at the origin this is $x^2 + y^2 + z^2 = r^2$. A point is inside the sphere if $CP < r$, on it if $CP = r$ and outside it if $CP > r$.

The general form

Expand the equation and collect the terms:

Compare with $x^2 - 2ax + a^2 + \ldots$. So $u = -a$, $v = -b$, $w = -c$. The centre is $(-u, -v, -w)$ and $r^2 = u^2 + v^2 + w^2 - d$. This is a real sphere only if $r^2 > 0$.

Three things mark an equation as a sphere: the squares have the same coefficient $1$, there are no products like $xy$, and $u^2 + v^2 + w^2 - d$ is positive.

A worked example

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

1. Read the coefficients. $2u = -4$, $2v = 6$, $2w = -2$ and $d = -11$. So $u = -2$, $v = 3$ and $w = -1$.
2. Centre. $(-u, -v, -w) = (2, -3, 1)$.
3. Radius. $r^2 = 4 + 9 + 1 + 11 = 25$, so $r = 5$.

A sphere with a given diameter

If $A$ and $B$ are the ends of a diameter, the centre is the midpoint of $AB$ and the radius is $\dfrac{AB}{2}$. The equation can also be written as

For $A(1, 2, 3)$ and $B(5, -2, 7)$, the centre is $(3, 0, 5)$ and $AB^2 = 16 + 16 + 16 = 48$. So $r^2 = 12$ and the equation is $(x - 3)^2 + y^2 + (z - 5)^2 = 12$.

Where a sphere meets the planes and the axes

With a plane. The plane $z = h$ cuts a sphere of centre $(a, b, c)$ and radius $r$ in a circle. Its centre is $(a, b, h)$ and its radius is $\rho = \sqrt{r^2 - (h - c)^2}$. This comes from Pythagoras in a right triangle with the radius $r$ of the sphere as hypotenuse. There is a circle only if $|h - c| < r$.

With an axis. Put $y = 0$ and $z = 0$ in the equation. The sphere meets the $x$-axis where $(x - a)^2 = r^2 - b^2 - c^2$. The chord on the $x$-axis has length $2\sqrt{r^2 - b^2 - c^2}$.

For the sphere of the worked example (centre $(2, -3, 1)$ and $r = 5$): on the $x$-axis $(x - 2)^2 = 25 - 9 - 1 = 15$, so the chord has length $2\sqrt{15}$. The plane $z = 0$ is $1$ unit from the centre, so it cuts a circle of radius $\sqrt{24} = 2\sqrt{6}$.

A sphere through the origin. Its equation has $d = 0$. A sphere that touches a coordinate plane has a radius equal to the distance of its centre from that plane. For example, a sphere with centre $(3, -4, 5)$ that touches the $xy$-plane has radius $5$.

Practise with new numbers

Try it in Python

The function centre_and_radius turns the coefficients into a centre and a radius. The program then finds where the sphere of the worked example meets the $x$-axis and the $xy$-plane.

Common mistakes

• Reading the centre with the wrong signs. In $(x - a)^2$ the centre has $x = a$. In $x^2 + 2ux$ it has $x = -u$.
• Taking half the coefficient wrongly. The coefficient of $x$ is $2u$, so $u$ is half of it.
• Forgetting the squares in the radius. $r^2 = u^2 + v^2 + w^2 - d$, and then the root.
• Calling any equation with squares a sphere. A term $xy$, unequal coefficients of the squares, or $r^2 \le 0$ means it is not a sphere.

Remember these

• Sphere: $(x - a)^2 + (y - b)^2 + (z - c)^2 = r^2$. General form: centre $(-u, -v, -w)$ and $r^2 = u^2 + v^2 + w^2 - d$.
• Ends of a diameter: the centre is the midpoint, the radius is half of $AB$.
• Plane $z = h$ cuts a circle of radius $\sqrt{r^2 - (h - c)^2}$. The chord on the $x$-axis is $2\sqrt{r^2 - b^2 - c^2}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Write the equation of the sphere with centre $(1, -2, 3)$ and radius $4$.

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

3. Is the point $(3, 2, -1)$ inside, on or outside the sphere $(x - 1)^2 + (y - 2)^2 + (z + 1)^2 = 4$?

4. Find the equation of the sphere that has $A(1, 2, 3)$ and $B(5, -2, 7)$ as the ends of a diameter. Check it with the worked example in the lesson.

5. A sphere has centre $(3, -4, 5)$ and touches the $xy$-plane. What is its radius, and what are the points of the $z$-axis on it?

References

• NCERT, Mathematics Textbook for Class XI, the chapter Introduction to Three Dimensional Geometry. The sphere is an extension that the Indian entrance exams use.
• OpenStax, Calculus Volume 3, Chapter 2, section 2.2 on vectors in three dimensions, the part on spheres.