# Question Video: Using the Given Equation of a Sphere to Find Its Centre and Radius Mathematics

Given that a sphere’s equation is (𝑥 + 5)² + (𝑦 − 12)² + (𝑧 − 2)² − 289 = 0, determine its center and radius.

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### Video Transcript

Given that a sphere’s equation is 𝑥 plus five all squared plus 𝑦 minus 12 all squared plus 𝑧 minus two all squared minus 289 is equal to zero, determine its center and radius.

In this question, we’re given an equation, and we’re told that this equation represents a sphere. We need to determine the center of this sphere and the radius of this sphere. To do this, we can look at the equation we’re given, and we can see it’s very similar to the standard form for the equation of a sphere. So let’s start by recalling the standard form for the equation of a sphere. We recall a sphere of radius 𝑟 centered at the point 𝑎, 𝑏, 𝑐 will have the following equation in standard form. 𝑥 minus 𝑎 all squared plus 𝑦 minus 𝑏 all squared plus 𝑧 minus 𝑐 all squared is equal to 𝑟 squared. Therefore, if we can rewrite the equation we’re given in the question into standard form, we’ll be able to find the center of our sphere and we’ll also be able to find the radius of our sphere.

So let’s start with the equation of the sphere given to us in the question. We can see in the standard form for the equation of our sphere, our constant is on the right-hand side of our equation. However, in our equation, we can see it’s on the left. So the first thing we’re going to need to do is add 289 to both sides of our equation. This gives us 𝑥 plus five all squared plus 𝑦 minus 12 all squared plus 𝑧 minus two all squared is equal to 289. And now we can see our equation is almost in the correct form. However, because we want to find the radius of this sphere, we’re going to write our constant on the right-hand side as a square. And by taking the square root of 289, we can see that 289 is equal to 17 squared. So we can just rewrite the right-hand side of this equation as 17 squared.

And now our equation is almost exactly in standard form. However, we can see in the standard form, we need to subtract the constants inside of our parentheses. However, on the left-hand side of our equation, inside our parentheses, we have 𝑥 plus five. And we can fix this by realizing 𝑥 plus five is the same as 𝑥 minus negative five. So we’ll rewrite this term as 𝑥 minus negative five all squared. And now that we’ve written this equation in the standard form for the equation of a sphere, we can find the center and radius of this sphere. The center of our sphere will be the point negative five, 12, two because these are the values of 𝑎, 𝑏, and 𝑐 in our standard form for the equation of a sphere. And another way of thinking about this is these are the values of 𝑥, 𝑦, and 𝑧 which will make each term on the left-hand side of our equation equal to zero. And we can also find the radius of this sphere. The radius of this sphere is going to be equal to 17.

And remember, the radius of a sphere represents a length, so we can give this units. We’ll call this 17 length units. Therefore, given the equation of the sphere is 𝑥 plus five all squared plus 𝑦 minus 12 all squared plus 𝑧 minus two all squared minus 289 is equal to zero, by rewriting this equation into the standard form for the equation of a sphere, we were able to find its center and radius. We were able to show the center of the sphere was the point negative five, 12, two and the radius of this sphere was 17 length units.