Video: Finding the Equation of a Sphere given Its Radius and the Coordinates of Its Centre

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

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Video Transcript

Give the equation of the sphere of center 11, eight, negative five and radius three in standard form.

Let’s recall first of all what is meant by the standard form of the equation of a sphere. If a sphere has its center at the point with coordinates π‘Ž, 𝑏, 𝑐 and a radius of π‘Ÿ units, then the equation of that sphere in standard form is given by π‘₯ minus π‘Ž squared plus 𝑦 minus 𝑏 squared plus 𝑧 minus 𝑐 squared is equal to π‘Ÿ squared, where π‘₯, 𝑦, 𝑧 are the coordinates of any point on the surface of the sphere.

We’ve been given both the center and radius of our sphere. So we can substitute the values of π‘Ž, 𝑏, 𝑐, and π‘Ÿ into the standard form. We get π‘₯ minus 11 squared plus 𝑦 minus eight squared plus 𝑧 minus negative five squared is equal to three squared.

Be careful here. A common mistake would be to forget one of the minus signs and write down just 𝑧 minus five squared. But we’re subtracting negative five. So we must write 𝑧 minus negative five squared. We can of course simplify this though because the two negative signs together will form a positive. 𝑧 minus negative five is equal to 𝑧 plus five. So we can replace that third bracket with 𝑧 plus five squared. On the right-hand side of our equation, three squared is equal to nine.

So the equation of the sphere of center 11, eight, negative five and radius three in standard form is π‘₯ minus 11 squared plus 𝑦 minus eight squared plus 𝑧 plus five squared is equal to nine. If we were asked to give the equation in expanded form, we’d need to expand each of the brackets and then simplify the result by collecting like terms.

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