Lesson Explainer: Equation of a Sphere Mathematics

In this explainer, we will learn how to find the equation of a sphere given its center and how to find the center and the radius given the sphere’s equation.

Let us start by recalling the definition of a sphere.

Definition: Sphere

A sphere is the locus of all points (π‘₯,𝑦,𝑧) that are a distance π‘Ÿ from a fixed point (π‘Ž,𝑏,𝑐).

In this definition, π‘Ÿ is the radius of the sphere, and the fixed point (π‘Ž,𝑏,𝑐) is the center of the sphere. We now want to use this definition to help us derive the equation of a sphere in standard form.

First, recall that we can calculate the distance between two points (π‘₯,𝑦,𝑧) and (π‘₯,𝑦,𝑧) using the formula 𝑑=(π‘₯βˆ’π‘₯)+(π‘¦βˆ’π‘¦)+(π‘§βˆ’π‘§).

In our definition of a sphere, we have a set of points (π‘₯,𝑦,𝑧) that are a fixed distance π‘Ÿ from the center of the sphere (π‘Ž,𝑏,𝑐). Therefore, by substituting into the distance formula, we have that π‘Ÿ=(π‘₯βˆ’π‘Ž)+(π‘¦βˆ’π‘)+(π‘§βˆ’π‘), which is an equation for the sphere. This, however, is not the standard form for the equation of a sphere. If we square both sides of the equation, we get π‘Ÿ=(π‘₯βˆ’π‘Ž)+(π‘¦βˆ’π‘)+(π‘§βˆ’π‘), which is the standard form of the equation of a sphere with radius π‘Ÿ (noting that π‘Ÿ>0) and center (π‘Ž,𝑏,𝑐).

Definition: The Equation of a Sphere in Standard Form

The Cartesian equation of a sphere with radius π‘Ÿ and center (π‘Ž,𝑏,𝑐), in standard form, is (π‘₯βˆ’π‘Ž)+(π‘¦βˆ’π‘)+(π‘§βˆ’π‘)=π‘Ÿ.

This means that if we know the coordinates for the center of a sphere along with the length of its radius, or if we can calculate these two pieces of information, then we can find the equation of the sphere.

With that in mind, let us look at our first couple of examples.

Example 1: Finding the Equation of a Sphere given Its Center and Radius

Give the equation of the sphere of center (11,8,βˆ’5) and radius 3 in standard form.

Answer

We know that the standard form of the equation of a sphere is (π‘₯βˆ’π‘Ž)+(π‘¦βˆ’π‘)+(π‘§βˆ’π‘)=π‘Ÿ, where (π‘Ž,𝑏,𝑐) is the center and π‘Ÿ is the length of the radius. Here, we are given the coordinates of the center of the sphere and, therefore, can deduce that π‘Ž=11, 𝑏=8, and 𝑐=βˆ’5. We are also told that π‘Ÿ=3. Substituting these values, we find that (π‘₯βˆ’11)+(π‘¦βˆ’8)+(𝑧+5)=3.

Finally, evaluating the right-hand side gives (π‘₯βˆ’11)+(π‘¦βˆ’8)+(𝑧+5)=9.

Example 2: Finding the Center and Radius of a Sphere given Its Equation

Given that a sphere’s equation is (π‘₯+5)+(π‘¦βˆ’12)+(π‘§βˆ’2)βˆ’289=0, determine its center and radius.

Answer

We know that the standard form of the equation of a sphere is (π‘₯βˆ’π‘Ž)+(π‘¦βˆ’π‘)+(π‘§βˆ’π‘)=π‘Ÿ, where (π‘Ž,𝑏,𝑐) is the center and π‘Ÿ is the length of the radius. We need to rearrange the given equation into this form. First, we add 289 to both sides of the equation to give us (π‘₯+5)+(π‘¦βˆ’12)+(π‘§βˆ’2)=289.

Then, we rewrite the expression in the first set of parentheses to match the standard form equation to get (π‘₯βˆ’(βˆ’5))+(π‘¦βˆ’12)+(π‘§βˆ’2)=289.

Finally, noting that the square root of 289=17, we can rewrite our equation once more to get (π‘₯βˆ’(βˆ’5))+(π‘¦βˆ’12)+(π‘§βˆ’2)=17.

From here, we can determine that the center of the sphere has coordinates (βˆ’5,12,2) and that the radius is 17.

Let us now look at an example where we apply what we have learned about the equation of a sphere to solve a geometric problem.

Example 3: Finding the Endpoint of a Sphere’s Diameter given the Other Endpoint and the Sphere’s Equation

Given 𝐴(0,4,4) and that 𝐴𝐡 is a diameter of the sphere (π‘₯+2)+(𝑦+1)+(π‘§βˆ’1)=38, what is the point 𝐡?

Answer

We can take two approaches to solve this problem. We can either solve the problem geometrically using vectors or solve it algebraically by considering the relative positions of the points that we know.

First, recall the standard form for the equation of a sphere: (π‘₯βˆ’π‘Ž)+(π‘¦βˆ’π‘)+(π‘§βˆ’π‘)=π‘Ÿ, where (π‘Ž,𝑏,𝑐) is the center and π‘Ÿ is the length of the radius. Using this, we can see that the center of our sphere has coordinates (βˆ’2,βˆ’1,1) and we can draw a quick sketch of the sphere to help us visualize the problem.

We note that the points 𝐴, 𝐢, and 𝐡 lie on the same line, and because 𝐴𝐡 is a diameter of the sphere, we know that 𝐢 is the midpoint of 𝐴𝐡.

Method 1

First, let us look at how to solve the problem algebraically. As 𝐴 and 𝐡 are the end points of a diameter of the sphere and 𝐢 is the center, we know that 𝐢 is the midpoint of 𝐴𝐡. For a line in three-dimensional space, the midpoint, 𝐢, of a line segment with end points 𝐴(π‘₯,𝑦,𝑧) and 𝐡(π‘₯,𝑦,𝑧) can be calculated using the formula 𝐢=ο€Όπ‘₯+π‘₯2,𝑦+𝑦2,𝑧+𝑧2.

We can then deduce that ο€Ό0+π‘₯2,4+𝑦2,4+𝑧2=(βˆ’2,βˆ’1,1).

This gives us three equations: π‘₯2=βˆ’2,4+𝑦2=βˆ’1,4+𝑧2=1.

Solving these gives us π‘₯=βˆ’4, 𝑦=βˆ’6, and 𝑧=βˆ’2. Therefore, the coordinates of 𝐡 are (βˆ’4,βˆ’6,βˆ’2).

Method 2

For interest, if we want to solve the problem geometrically using vectors, we can deduce that 𝐴𝐢=οƒŸπΆπ΅; therefore, given that point 𝑂 is a relative center of the system, we have that οƒŸπ‘‚π΅=𝑂𝐢+𝐴𝐢.

We have that 𝑂𝐢=(βˆ’2,βˆ’1,1) and 𝐴𝐢=οƒ π‘‚πΆβˆ’οƒ π‘‚π΄. Therefore, 𝐴𝐢=(βˆ’2,βˆ’1,1)βˆ’(0,4,4), which simplifies to (βˆ’2,βˆ’5,βˆ’3).

We can now calculate οƒŸπ‘‚π΅ as follows: οƒŸπ‘‚π΅=(βˆ’2,βˆ’1,1)+(βˆ’2,βˆ’5,βˆ’3), which simplifies to (βˆ’4,βˆ’6,βˆ’2).

Therefore, the coordinates of 𝐡 are (βˆ’4,βˆ’6,βˆ’2).

Before we look at our final example, let us recall how to complete the square for a quadratic expression.

How To: Completing the Square for a Quadratic Expression

Let us consider the expression π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨.

To complete the square, we start by factoring out the π‘Ž: π‘Žο€½π‘₯+π‘π‘Žπ‘₯+π‘π‘Žο‰.

Now, we consider the expression π‘Žο€½π‘₯+𝑏2π‘Žο‰οŠ¨. If we expand the square of the binomial, we find that π‘Žο€½π‘₯+𝑏2π‘Žο‰=π‘Žο€Ώπ‘₯+π‘π‘Žπ‘₯+𝑏(2π‘Ž), which we note is the same as the expression above, excluding the constant term.

Therefore, we can rewrite our original expression as follows: π‘Žο€Ώο€½π‘₯+𝑏2π‘Žο‰βˆ’π‘(2π‘Ž)+π‘π‘Žο‹.

Finally, multiplying through by π‘Ž gives π‘Žο€½π‘₯+𝑏2π‘Žο‰βˆ’π‘4π‘Ž+𝑐.

This is the form of our original expression after completing the square.

We can use the method outlined above to convert the general form of the equation of a sphere to standard form. If we consider the sphere whose general equation is π‘₯+𝑦+π‘§βˆ’2π‘Žπ‘₯βˆ’2π‘π‘¦βˆ’2𝑐𝑧+𝑑=0, then we can covert this to standard form by completing the squares for the quadratics in each of the three variables. If we start by rearranging the equation to group all of the terms containing the same variables, we get π‘₯βˆ’2π‘Žπ‘₯+π‘¦βˆ’2𝑏𝑦+π‘§βˆ’2𝑐𝑧+𝑑=0.

Now, we can complete the square for each of the three quadratics to get (π‘₯βˆ’π‘Ž)βˆ’π‘Ž+(π‘¦βˆ’π‘)βˆ’π‘+(π‘§βˆ’π‘)βˆ’π‘+𝑑=0.

Finally, if we group together the constant terms, we can see that π‘Ÿ=π‘Ž+𝑏+π‘βˆ’π‘‘, and the center of the sphere has coordinates (π‘Ž,𝑏,𝑐).

To finish, let us look at an example where we need to identify the center and radius of a sphere whose equation is given in general form.

Example 4: Identifying the Center and Radius of a Sphere given Its Equation in General Form

Give the center and radius of the sphere π‘₯+𝑦+π‘§βˆ’8π‘₯+8𝑦+10𝑧+8=0.

Answer

Here, we are given the general form of the equation of a sphere, but we will need to compare the equation with the standard form of the equation of a sphere to determine its center and radius. Recall that the standard form for the equation of a sphere is (π‘₯βˆ’π‘Ž)+(π‘¦βˆ’π‘)+(π‘§βˆ’π‘)=π‘Ÿ, where (π‘Ž,𝑏,𝑐) is the center and π‘Ÿ is the length of the radius.

First, let us group together the terms that contain the same variables: π‘₯βˆ’8π‘₯+𝑦+8𝑦+𝑧+10𝑧+8=0.

Now, we need to complete the square for the quadratics in each of the three variables, giving us (π‘₯βˆ’4)βˆ’16+(𝑦+4)βˆ’16+(𝑧+5)βˆ’25+8=0.

Collecting the constant terms gives (π‘₯βˆ’4)+(𝑦+4)+(𝑧+5)βˆ’49=0, and adding 49 to each side gives us (π‘₯βˆ’4)+(𝑦+4)+(𝑧+5)=49.

Finally, if we compare this with the standard form for the equation of a sphere, we can see that the center of the sphere is (4,βˆ’4,βˆ’5) and the radius of the sphere is √49=7.

Key Points

  • A sphere is a three-dimensional shape where every point is a distance π‘Ÿ (the radius of the sphere) from the center.
  • A sphere centered at the point (π‘Ž,𝑏,𝑐) with radius π‘Ÿ has the equation (in standard form) (π‘₯βˆ’π‘Ž)+(π‘¦βˆ’π‘)+(π‘§βˆ’π‘)=π‘Ÿ. If we are given the equation of the sphere not in standard form, we can convert it to this form using algebraic methods and identify the center and radius by comparing the resulting equation to the general form.
  • To identify the center and radius of a sphere whose equation is given in general form, we can convert the equation to standard form by completing the square for each of the quadratics in each of the three variables.

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