Video: Finding the Equation of a Sphere given the Coordinates of Its Center and That of a Point It Passes through

Which of the following is the equation of the sphere of center (8, βˆ’15, 10) and passing through (βˆ’14, 13, βˆ’14)? [A] (π‘₯ βˆ’ 8)Β² + (𝑦 + 15)Β² + (𝑧 βˆ’ 10)Β² = 1,844 [B] (π‘₯ βˆ’ 8)Β² βˆ’ (𝑦 + 15)Β² βˆ’ (𝑧 βˆ’ 10)Β² = 1,844 [C] (π‘₯ βˆ’ 8)Β² + (𝑦 + 15)Β² + (𝑧 βˆ’ 10)Β² = 56 [D] (π‘₯ βˆ’ 8)Β² βˆ’ (𝑦 + 15)Β² βˆ’ (𝑧 βˆ’ 10)Β² = 56

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Video Transcript

Which of the following is the equation of the sphere of center eight, negative 15, 10 and passing through the point negative 14, 13, negative 14? Is it option (A) π‘₯ minus eight all squared plus 𝑦 plus 15 all squared plus 𝑧 minus 10 all squared is equal to 1,844? Is it option (B) π‘₯ minus eight all squared minus 𝑦 plus 15 all squared minus 𝑧 minus 10 all squared is equal to 1,844? Or is it option (C) π‘₯ minus eight all squared plus 𝑦 plus 15 all squared plus 𝑧 minus 10 all squared is equal to 56? Or finally, is it option (D) π‘₯ minus eight all squared minus 𝑦 plus 15 all squared minus 𝑧 minus 10 all squared is equal to 56?

In this question, we’re asked to determine which of four equations represents the equation of a sphere. We’re told that the center of the sphere is the point eight, negative 15, 10. And we’re also told that this sphere passes through the point negative 14, 13, negative 14. There’re several different methods we can use to answer this question. For example, we could look at the options given to us and try to eliminate some of these options. However, by looking at these options, we can see they’re very similar to the standard form of the equation of a sphere. So we’ll answer this question by just finding the standard form of the equation of a sphere with center eight, negative 15, 10 passing through the point negative 14, 13, negative 14.

We recall a sphere centered at the point π‘Ž, 𝑏, 𝑐 with a radius of π‘Ÿ, which must be positive, has 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. This is called the standard form of the equation of the sphere. All spheres can be represented in standard form. All we need to know to find the standard form of the equation of a sphere is its center and its radius. In this question, we’re already told that the center of the sphere is the point eight, negative 15, 10. Therefore, in the standard form of the equation of the sphere, the value of π‘Ž will be eight, the value of 𝑏 will be negative 15, and the value of 𝑐 will be 10.

We can substitute these values into the standard form of the equation of the sphere. This means that the standard form of the equation of the sphere given to us in the question must be in the form π‘₯ minus eight all squared plus 𝑦 plus 15 all squared plus 𝑧 minus 10 all squared is equal to π‘Ÿ squared, where π‘Ÿ will be the radius of the sphere. All we need to find now to find the standard form of the equation of the sphere is to find the value of the radius π‘Ÿ. To find this radius, we need to recall exactly what we mean by the radius of a sphere.

Remember, a sphere is a shape in three dimensions which represents all points equidistant from a point called the center. This distance is called the radius of the sphere. And since we know that our sphere passes through the point negative 14, 13, negative 14 and we know the coordinates of the center of the sphere, the distance between these two points must be equal to the radius of the sphere. Therefore, we can find the radius of this sphere by finding the distance between these two points.

And to do this, we recall the formula for finding the distance between two points in three-dimensional space. The distance between the point π‘₯ sub one, 𝑦 sub one, 𝑧 sub one and the point π‘₯ sub two, 𝑦 sub two, 𝑧 sub two is given by the square root of π‘₯ sub one minus π‘₯ sub two all squared plus 𝑦 sub one minus 𝑦 sub two all squared plus 𝑧 sub one minus 𝑧 sub two all squared. And it doesn’t matter in which order we label the two points; the distance between them will be the same.

Substituting the point on our sphere and the center of our sphere into the formula for the distance between two points in three-dimensional space gives us that the radius of the sphere is equal to the square root of negative 14 minus eight all squared plus 13 minus negative 15 all squared plus negative 14 minus 10 all squared. Simplifying the expression inside of our square root symbol, we get that π‘Ÿ is equal to the square root of 1,844.

Now we could simplify this expression further; however, remember, we’re trying to find an expression for π‘Ÿ squared for the standard form of the equation of our sphere. This means we’re going to want to square both sides of this equation anyway. Doing this gives us that π‘Ÿ squared will be equal to 1,844. Finally, we can substitute this value of π‘Ÿ squared into the standard form of the equation of our sphere.

This gives us that the standard form of the equation of a sphere centered at the point eight, negative 15, 10 passing through the point negative 14, 13, negative 14 π‘₯ minus eight all squared plus 𝑦 plus 15 all squared plus 𝑧 minus 10 all squared is equal to 1,844, which we can see is given by answer (A).

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