Question Video: Finding the Center and Radius of a Sphere Given Its Equation Mathematics

Give the center and radius of the sphere π‘₯Β² + 𝑦² + 𝑧² + 7π‘₯ + 6𝑦 + 3𝑧 + 12 = 0.

05:25

Video Transcript

Give the center and radius of the sphere π‘₯ squared plus 𝑦 squared plus 𝑧 squared plus seven π‘₯ plus six 𝑦 plus three 𝑧 plus 12 is equal to zero.

In this question, we’re given an equation which represents a sphere. We need to find the center of this sphere and the radius of this sphere. To answer this question, we first notice the equation of our sphere that we’re given is in the general form for the equation of a sphere. However, it would be easier if we had this written in the standard form for the equation of a sphere.

We recall the standard form for the equation of a sphere tells us a sphere with center the point π‘Ž, 𝑏, 𝑐 and a radius of π‘Ÿ will have the equation π‘₯ minus π‘Ž all squared plus 𝑦 minus 𝑏 all squared plus 𝑧 minus 𝑐 all squared is equal to π‘Ÿ squared. The equation of any sphere can be represented in this form. This means we can rewrite the equation given to us in this question in this form to find its center and its radius.

To rewrite this equation in this form, let’s start by looking at the π‘₯-terms on the left-hand side of our equation. We need to write this in the form π‘₯ minus π‘Ž all squared. We can do this by completing the square. We need to take one-half of the coefficient of our π‘₯-term. This gives us seven over two.

We can now consider π‘₯ plus seven over two all squared. If we were to evaluate this either by using binomial expansion or the FOIL method, we would get π‘₯ squared plus seven π‘₯ plus 49 over four. We can see this is almost exactly equal to the expression we want: π‘₯ squared plus seven π‘₯. All we’ve done is added an extra constant of 49 over four. Therefore, if we subtract 49 over four from π‘₯ plus seven over two all squared, we would get π‘₯ squared plus seven π‘₯.

Therefore, by completing the square, we’ve shown π‘₯ squared plus seven π‘₯ is equal to π‘₯ plus seven over two all squared minus 49 over four. We can then substitute this into our equation of our sphere for π‘₯ squared plus seven π‘₯. However, let’s first find an expression for our 𝑦-terms, 𝑦 squared plus six 𝑦, once again by completing the square.

Once again, we’re going to need to half the coefficient of 𝑦. That’s six over two, which is equal to three. This gives us 𝑦 plus three all squared. However, if we were to distribute the exponent over our parentheses, we would get 𝑦 squared plus six 𝑦 plus nine. Therefore, if we subtract nine from this expression, we would get that it’s equal to 𝑦 squared plus six 𝑦.

Finally, we’re going to need to do the same with our 𝑧-terms. In the equation of the sphere given to us, the 𝑧-terms are 𝑧 squared plus three 𝑧. Once again, we’re going to need to complete the square. We need to divide the coefficient of 𝑧 by two. In this case, that gives us three over two. This means that 𝑧 plus three over two all squared will be equal to 𝑧 squared plus three 𝑧. But then we add an extra constant of nine over four. Therefore, we’re going to need to subtract this extra constant of nine over four to make this equal to 𝑧 squared plus three 𝑧.

Now we can substitute all three of these expressions we found by completing the square into the equation of our sphere. We’ll start by replacing both of the π‘₯-terms of π‘₯ plus seven over two all squared minus 49 over four. Then we can replace both of the 𝑦-terms with 𝑦 plus three all squared minus nine. Then we can replace the two 𝑧-terms in this equation with 𝑧 plus three over two all squared minus nine over four. Finally, the equation of our sphere tells us if we add 12 to this expression, then it must be equal to zero.

This is now almost the standard form for the equation of our sphere. We just need our constant on the right-hand side of our equation and to be in the form π‘Ÿ squared. To write our equation in this form, we need to calculate negative 49 over four minus nine minus nine over four plus 12. This gives us negative 23 over two. So we’ll add 23 over two to both sides of this equation. Doing this and simplifying, we get π‘₯ plus seven over two all squared plus 𝑦 plus three all squared plus 𝑧 plus three over two all squared will be equal to 23 over two.

This is now almost in the standard form of the equation of a sphere. However, the constant on the right-hand side of our equation is usually written in the form π‘Ÿ squared. This is because if π‘Ÿ is positive, this would then be the radius of our sphere. One way of doing this is just to take the positive square root of this number because, of course, the positive square root of this number all squared won’t change its value. It will still be equal to 23 over two. This means the radius of this sphere is the square root of 23 over two.

However, we can simplify this. First, let’s simplify our radical by taking the square root of 23 and then dividing this by the square root of two. We can then rationalize the denominator by multiplying both the numerator and denominator by the square root of two. Evaluating the numerator, we get root 23 multiplied by root two will be the square root of 23 times two, which is the square root of 46. And in our denominator, we get root two multiplied by root two, which is just equal to two. This is another way we can represent the value of the radius of our sphere π‘Ÿ.

Finally, we need to find the center of this sphere. We can do this by solving each of our binomials equal to zero. Or alternatively we can just multiply the constant in the binomial by negative one. We get π‘Ž is negative seven over two, 𝑏 is negative three, and 𝑐 is negative three over two.

Therefore, we’ve shown the sphere given by the equation π‘₯ squared plus 𝑦 squared plus 𝑧 squared plus seven π‘₯ plus six 𝑦 plus three 𝑧 plus 12 is equal to zero has center negative seven over two, negative three, negative three over two and a radius of root 46 over two.

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