### Video Transcript

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

Letβs 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, 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.

Weβve been given both the center
and radius of our sphere. So we can substitute the values of
π, π, π, and π into the standard form. We get π₯ minus 11 squared plus π¦
minus eight squared plus π§ 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 π§ minus five squared. But weβre subtracting negative
five. So we must write π§ minus negative
five squared. We can of course simplify this
though because the two negative signs together will form a positive. π§ minus negative five is equal to
π§ plus five. So we can replace that third
bracket with π§ 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 π¦ minus eight squared plus π§ plus five squared is equal to nine. 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.