WEBVTT
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Give the equation of the sphere of center 11, eight, negative five and radius three in standard form.
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Letβs recall first of all what is meant by the standard form of the equation of a sphere.
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If a sphere has its center at the point with coordinates π, π, π and a radius of π units, then the equation of that sphere in standard form is given by π₯ minus π squared plus π¦ minus π squared plus π§ minus π squared is equal to π squared, where π₯, π¦, π§ are the coordinates of any point on the surface of the sphere.
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Weβve been given both the center and radius of our sphere.
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So we can substitute the values of π, π, π, and π into the standard form.
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We get π₯ minus 11 squared plus π¦ minus eight squared plus π§ minus negative five squared is equal to three squared.
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Be careful here.
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A common mistake would be to forget one of the minus signs and write down just π§ minus five squared.
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But weβre subtracting negative five.
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So we must write π§ minus negative five squared.
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We can of course simplify this though because the two negative signs together will form a positive.
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π§ minus negative five is equal to π§ plus five.
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So we can replace that third bracket with π§ plus five squared.
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On the right-hand side of our equation, three squared is equal to nine.
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So the equation of the sphere of center 11, eight, negative five and radius three in standard form is π₯ minus 11 squared plus π¦ minus eight squared plus π§ plus five squared is equal to nine.
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If we were asked to give the equation in expanded form, weβd need to expand each of the brackets and then simplify the result by collecting like terms.