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Equation of a Sphere
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In this lesson, weβre going to learn the standard form for the equation of a sphere.
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And weβre going to learn how to find the standard form for the equation of a sphere given the center of our sphere and the radius of our sphere.
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And weβll also see how to find the center and radius of our sphere given the equation of the sphere in standard form.
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So to find the equation of a sphere, weβre first going to need to recall exactly what we mean by a sphere.
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We recall a sphere centered at the point π with a radius of π will consist of all points a distance of π units away from the point π.
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And thereβs a lot of things weβre pointing out about this definition.
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First, spheres are in three dimensions.
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So, in particular, all of the points on our sphere will be given in three-dimensional coordinates and the point π will also be given in three-dimensional coordinates.
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And itβs important we remember this because if we instead use two dimensions, this would just be the definition of a circle.
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Next, since the radius π represents a length or a distance, we say that π must be positive.
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If our radius π was equal to zero, then our sphere would just be all points at a distance of zero from π.
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It would just be the point π itself.
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And if π was negative, our definition wouldnβt make sense.
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So we just said π has to be positive.
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So now that we have the definition of our sphere, we want to find the equation of our sphere.
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And thereβs a few steps to doing this.
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Remember, for the equation for our sphere to be correct, every point on our sphere must satisfy our equation, and every point which satisfies our equation must lie on the sphere.
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And we want to find the equation for a general sphere.
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So weβll set the center of our sphere to be the point π, π, π and weβll let π, the point π₯, π¦, π§, lie on our sphere of radius π centered at the point π.
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And this is the general sphere weβre going to try and find the equation of.
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One way of finding the equation of this sphere would be to copy the method we use to find the equation of a circle.
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We drew a picture of our circle and then used what we knew about the Pythagorean theorem to find an equation for our circle.
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In fact, this method would work in exactly the same way.
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It would be slightly more complicated because we would be working in three dimensions.
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However, this would give us the correct equation for a sphere.
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However, we actually have more powerful tools to deal with this, giving us an easier method to find the equation of the sphere.
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Remember, every point on our sphere must be a distance of π from our point π.
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In our case, the point π must be a distance π from the center of our sphere π.
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And we know a formula to calculate the distance between two points in three dimensions.
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We recall, to calculate the distance between the point π₯ one, π¦ one, π§ one and the point π₯ two, π¦ two, π§ two, we just calculate the square root of π₯ two minus π₯ one all squared plus π¦ two minus π¦ one all squared plus π§ two minus π§ one all squared.
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So letβs use this formula to calculate the distance between point π and the point π.
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Remember, we already know that this distance is going to be equal to π.
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This gives us the distance between the point π and the point π is the square root of π₯ minus π all squared plus π¦ minus π all squared plus π§ minus π all squared.
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And we know this is going to be equal to the radius of our sphere π.
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So letβs think about exactly what weβve just shown.
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Weβve shown any point π which lies on the sphere of radius π centered at our point π must satisfy this equation.
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And in fact, we can show the opposite is true as well.
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We can think about what would happen if the point π₯ zero, π¦ zero, π§ zero satisfied this equation.
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This means we substitute π₯ equals π₯ zero, π¦ equals π¦ zero, and π§ equals π§ zero into this equation.
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And the left-hand side of this equation will end up being equal to π.
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However, we can also see on the left-hand side of this equation, weβre just calculating the distance between the point π₯ zero, π¦ zero, π§ zero and the center of our sphere π.
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Therefore, if the point satisfies this equation, it must be a distance π from the center of our sphere π, which means it lies on our sphere.
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Therefore, weβve managed to find the equation of a sphere centered at the point π, π, π with a radius of π.
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Weβve shown every point on this sphere must satisfy this equation and every point which satisfies this equation must lie on our sphere.
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However, thereβs one thing you might have noticed.
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This is a very complicated-looking expression.
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In fact, we can simplify this by squaring both sides.
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And there is one thing worth pointing out here.
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We can justify that we wonβt be gaining extra solutions in this case because we know our value of π is positive.
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And everything inside of our square root symbol on the left-hand side of the equation is greater than or equal to zero because weβre taking the square of each term.
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This means we can guarantee squaring both sides will not give us extra solutions.
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And this gives us the following equation for our sphere.
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We call this the standard form for the equation of a sphere, just like we might call this equation the standard form for the equation of a circle if we did not have the third term on the left-hand side.
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So letβs clear some space and think about exactly what weβve just shown.
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We have shown the standard form for the equation of a sphere centered at the point π, π, π with a radius of π is given by π₯ minus π all squared plus π¦ minus π all squared plus π§ minus π all squared is equal to π squared.
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So now, given the center of our sphere and the radius of our sphere, we can find the equation of our sphere.
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And equally, if we were given the standard form for the equation of our sphere, we would be able to find the center of the sphere and the radius of our sphere.
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Letβs now see a few examples of how we can apply this.
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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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In this question, weβre asked to find the equation of a sphere.
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And not only this, weβre asked to find this equation in standard form.
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Weβre also given some information about our sphere.
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Weβre told the center of our sphere is 11, eight, negative five and the radius of our sphere is three.
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To answer this question, letβs start by recalling what we mean by the standard form of the equation of a sphere.
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We recall a sphere centered at the point π, π, π with a radius of π will have the following equation in standard form: π₯ minus π all squared plus π¦ minus π all squared plus π§ minus π all squared is equal to π squared.
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In the question, weβre told the center of our sphere is the point 11, eight, negative five, and weβre told the radius of our sphere is three.
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So all we need to do to answer this question is substitute these values into the standard form for the equation of a sphere.
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Substituting π is equal to 11, π is equal to eight, π is equal to negative five, and π is equal to three into our equation for a sphere, we get π₯ minus 11 all squared plus π¦ minus eight all squared plus π§ minus negative five all square is equal to three squared.
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And of course we can simplify this.
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π§ minus negative five all squared is equal to π§ plus five all squared, and three squared is equal to nine.
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And this gives us our final answer.
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The standard form for the equation of a sphere centered at the point 11, eight, negative five with a radius of three is given by π₯ minus 11 all squared plus π¦ minus eight all squared plus π§ plus five all squared is equal to nine.
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Letβs now see an example where weβre given the equation of a sphere and we need to find the sphereβs center and radius.
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Given that a sphereβs equation is π₯ plus five all squared plus π¦ minus 12 all squared plus π§ minus two all squared minus 289 is equal to zero, determine its center and radius.
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In this question, weβre given an equation, and weβre told that this equation represents a sphere.
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We need to determine the center of this sphere and the radius of this sphere.
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To do this, we can look at the equation weβre given, and we can see itβs very similar to the standard form for the equation of a sphere.
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So letβs start by recalling the standard form for the equation of a sphere.
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We recall a sphere of radius π centered at the point π, π, π will have the following equation in standard form.
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π₯ minus π all squared plus π¦ minus π all squared plus π§ minus π all squared is equal to π squared.
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Therefore, if we can rewrite the equation weβre given in the question into standard form, weβll be able to find the center of our sphere and weβll also be able to find the radius of our sphere.
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So letβs start with the equation of the sphere given to us in the question.
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We can see in the standard form for the equation of our sphere, our constant is on the right-hand side of our equation.
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However, in our equation, we can see itβs on the left.
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So the first thing weβre going to need to do is add 289 to both sides of our equation.
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This gives us π₯ plus five all squared plus π¦ minus 12 all squared plus π§ minus two all squared is equal to 289.
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And now we can see our equation is almost in the correct form.
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However, because we want to find the radius of this sphere, weβre going to write our constant on the right-hand side as a square.
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And by taking the square root of 289, we can see that 289 is equal to 17 squared.
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So we can just rewrite the right-hand side of this equation as 17 squared.
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And now our equation is almost exactly in standard form.
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However, we can see in the standard form, we need to subtract the constants inside of our parentheses.
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However, on the left-hand side of our equation, inside our parentheses, we have π₯ plus five.
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And we can fix this by realizing π₯ plus five is the same as π₯ minus negative five.
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So weβll rewrite this term as π₯ minus negative five all squared.
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And now that weβve written this equation in the standard form for the equation of a sphere, we can find the center and radius of this sphere.
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The center of our sphere will be the point negative five, 12, two because these are the values of π, π, and π in our standard form for the equation of a sphere.
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And another way of thinking about this is these are the values of π₯, π¦, and π§ which will make each term on the left-hand side of our equation equal to zero.
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And we can also find the radius of this sphere.
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The radius of this sphere is going to be equal to 17.
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And remember, the radius of a sphere represents a length, so we can give this units.
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Weβll call this 17 length units.
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Therefore, given the equation of the sphere is π₯ plus five all squared plus π¦ minus 12 all squared plus π§ 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.
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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.
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Letβs now see an example where we check whether a given equation is the equation of a sphere.
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Determine if the given equation π₯ squared plus π¦ squared plus π§ squared plus two π₯ minus two π¦ minus eight π§ plus 19 is equal to zero describes a sphere.
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If so, find its radius and center.
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In this question, weβre given an equation.
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We need to determine whether this equation represents a sphere.
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And if it does represent a sphere, we need to determine the center of this sphere and the radius of this sphere.
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The easiest way to do this would be to try and write our equation in the standard form for the equation of a sphere.
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So letβs start by recalling what this means.
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We recall a sphere centered at the point π, π, π with a radius π will have the following equation in standard form.
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π₯ minus π all squared plus π¦ minus π all squared plus π§ minus π all squared is equal to π squared.
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What we want to do is rewrite the equation given to us in the question as the equation of a sphere in standard form.
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And if we were able to do this, we would then be able to read off the center of our sphere and the radius of our sphere.
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To do this, weβre going to start by rearranging the equation given to us.
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Weβre going to write the π₯-terms first, then the π¦-terms, then the π§-terms.
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In the standard form for the equation of a sphere, we see we start with π₯ minus π all squared.
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However, in our equation, we have π₯ squared plus two π₯.
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To write π₯ squared plus two π₯ in this form, weβre going to need to complete the square.
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We recall to complete the square, we want to write these two terms in the form π₯ plus some constant all squared.
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And to find the constant, we need to halve the coefficient of π₯.
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This is because π₯ plus one all squared is equal to π₯ squared plus two π₯ plus one.
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If we want this to be equal to π₯ squared plus two π₯, we need to subtract one from both sides of our equation.
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Therefore, by completing the square, weβve shown that π₯ squared plus two π₯ is equal to π₯ plus one all squared minus one.
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Weβre then going to want to do exactly the same thing for our next two terms.
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This time, we halve the coefficient of π¦ to get negative one.
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And if we distribute the square over our parentheses, we get π¦ minus one all squared is equal to π¦ squared minus two π¦ plus one.
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And again, since we want this to be equal to π¦ squared minus two π¦, we need to subtract one from both sides of our equation, giving us that π¦ squared minus two π¦ is equal to π¦ minus one all squared minus one.
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Therefore, by completing the square, we were able to rewrite the two terms π¦ squared minus two π¦ as π¦ minus one all squared minus one.
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Finally, weβre going to need to complete the square on our π§-terms.
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Once again, inside our parentheses, weβre going to need to halve the coefficient of π§, which is negative eight.
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This gives us π§ minus four all squared.
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If we distribute the square over our parentheses, we get π§ squared minus eight π§ plus 16.
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So to make this equal to π§ squared minus eight π§, weβre going to need to subtract 16 from both sides of our equation.
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This gives us that π§ squared minus eight π§ is equal to π§ minus four all squared minus 16.
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So by completing the square, we were able to rewrite our term π§ squared minus eight π§ as π§ minus four all squared minus 16.
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And in our equation, we still need to add 19 and set this equal to zero.
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Therefore, by completing the square three times, we were able to rewrite the equation given to us in the question as π₯ plus one all squared minus one plus π¦ minus one all squared minus one plus π§ minus four all squared minus 16 plus 19 is equal to zero.
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And of course we can simplify this.
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We have negative one minus one minus 16 plus 19.
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If we evaluate this, we see itβs equal to one.
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This means we can rewrite the equation given to us in the question in the following form.
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But remember, in the standard form for the equation of a sphere, our constant is on the other side of the equation.
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So weβre going to subtract one from both sides of our equation.
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This gives us π₯ plus one all squared plus π¦ minus one all squared plus π§ minus four all squared is equal to negative one.
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But now, if we said that this was the equation for a sphere, we would have a problem.
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In this case, the radius would be the number which squares to give us negative one.
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However, our radius needs to be positive, so this doesnβt make any sense.
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So it doesnβt seem that this is the equation for a sphere.
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In fact, we can prove this.
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On the right-hand side of our equation, we can see we have a negative number.
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However, on the left-hand side of the equation, we can see all three of our terms are square.
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This means all three of our terms are greater than or equal to zero.
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So the left-hand side of our equation is greater than or equal to zero for all values of π₯, π¦, and π§.
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However, the right-hand side of our equation is negative.
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Therefore, not only is this not the equation of a sphere, there are no solutions to this equation at all.
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Therefore, to answer the question, does the equation π₯ squared plus π¦ squared plus π§ squared plus two π₯ minus two π¦ minus eight π§ plus 19 equals zero describe a sphere, we were able to show that no, this equation does not describe a sphere.
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Letβs now see an example where we use the equation of a sphere to find out some information geometrically about our sphere.
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Given that π΄ is the point zero, four, four and that the line segment π΄π΅ is a diameter of the sphere π₯ plus two all squared plus π¦ plus one all squared plus π§ minus one all squared is equal to 38, what is the point π΅?
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In this question, weβre given some information about a sphere.
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First, weβre told the standard equation of the sphere.
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Weβre also told that the line segment π΄π΅ is a diameter of our sphere and weβre told the coordinates at the point π΄.
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We need to use all of this information to determine the coordinates of point π΅.
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Thereβs several different methods we could do this.
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However, usually in problems like this, the easiest method involves starting by writing down all of the information weβre given.
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To do this, letβs start by recalling the standard form for the equation of a sphere.
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We recall a sphere of radius π centered at the point π, π, π will have the following equation in standard form.
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π₯ minus π all squared plus π¦ minus π all squared plus π§ minus π all squared is equal to π squared.
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This means if weβre given the equation of a sphere in standard form, we can find its center point π, π, π and we can also find its radius π.
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And we could see the equation given to us in the question is in standard form, so we can use this to find the center and radius of our sphere.
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Thereβs two different methods of finding the center point.
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We could rewrite the expressions inside of our parentheses as the variable minus some constant.
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However, we could also just find the value of the variable which makes this term equal to zero.
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So, for example, our value of π would be negative two, our value of π would be negative one, and our value of π would be one.
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Either method would work; itβs personal preference which one you want to use.
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Either way, weβve shown the center of the sphere given to us in the question is the point negative two, negative one, one.
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Similarly, we can find the radius of this sphere by taking the square root of 38.
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The last thing weβre going to want to do is use the fact that the line segment π΄π΅ is a diameter of our sphere and that the coordinates of the point π΄ are zero, four, four.
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And we might want to sketch this information on a sphere.
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However, itβs not necessary.
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We can actually just do this on a circle because if the line segment π΄π΅ is a diameter of the sphere, then itβs also a diameter of the circle of the same radius.
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In either case, the only information we need is the line segment π΄π΅ is a diameter of our sphere, so itβs a straight line passing through the center of our sphere.
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And we know the coordinates of the point π΄ is zero, four, four and the coordinates of our center π is negative two, negative one, one.
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And we can combine all of this information to find the coordinates of point π΅.
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First, line segment π΄πΆ and line segment πΆπ΅ are both radii of our sphere.
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Theyβre both going to have length π.
00:15:35.260 --> 00:15:42.180
Next, because we know the coordinates of the point π΄ and the coordinates of the point πΆ, weβre able to find the vector from π΄ to πΆ.
00:15:42.440 --> 00:15:43.990
And we can also see something interesting.
00:15:44.020 --> 00:15:50.730
This is going to be exactly the same as the vector from πΆ to π΅ because they have the same magnitude of π and they point in the same direction.
00:15:50.970 --> 00:15:52.850
So letβs use this to find the coordinates of π΅.
00:15:52.880 --> 00:15:54.920
First, weβre going to need to find the vector ππ.
00:15:54.920 --> 00:15:58.400
And to do this, we need to take the vector ππ and subtract the vector ππ.
00:15:58.640 --> 00:16:02.070
This gives us the following expression, and we can subtract this component-wise.
00:16:02.290 --> 00:16:06.560
This gives us the vector ππ is the vector negative two, negative five, negative three.
00:16:06.740 --> 00:16:08.330
We can then add this onto our sketch.
00:16:08.360 --> 00:16:12.390
And remember, the vector from πΆ to π΅ is also equal to the vector from π΄ to πΆ.
00:16:12.670 --> 00:16:17.130
Now we can find the coordinates of π΅ by adding our vector ππ to the vector ππ.
00:16:17.340 --> 00:16:22.230
And all weβre saying here is we can get to the point π΅ from the center by moving along the vector ππ.
00:16:22.450 --> 00:16:28.870
This gives us the vector ππ will be the vector negative two, negative one, one plus the vector negative two, negative five, negative three.
00:16:29.060 --> 00:16:34.350
And we add these together component-wise to get the vector ππ is the vector negative four, negative six, negative two.
00:16:34.600 --> 00:16:37.070
But remember, the question is not asking us for the vector ππ.
00:16:37.070 --> 00:16:39.100
Itβs asking us for the coordinates of the point π΅.
00:16:39.350 --> 00:16:43.230
And of course, the coordinates of the point π΅ will just be the components of our vector π.
00:16:43.230 --> 00:16:47.110
And this gives us that π΅ is the point negative four, negative six, negative two.
00:16:47.390 --> 00:17:02.670
Therefore, we were able to show if π΄ is the point zero, four, four and the line segment π΄π΅ is a diameter of the sphere π₯ plus two all squared plus π¦ plus one all squared plus π§ minus one all squared is equal to 38, then the point π΅ must have coordinates negative four, negative six, negative two.
00:17:05.040 --> 00:17:07.130
Letβs now go over the key points of this video.
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First, we know that a sphere is a three-dimensional shape where every point on our sphere is a set distance of π from the center.
00:17:13.450 --> 00:17:15.810
And we call this value of π the radius of our sphere.
00:17:16.180 --> 00:17:19.860
And much like in the case with a circle, we were able to find the equation of a sphere.
00:17:19.890 --> 00:17:25.350
We were able to show that a sphere centered at the point π, π, π, with a radius of π will have the following equation.
00:17:25.380 --> 00:17:28.440
And we call this equation the standard form of the equation of our sphere.
00:17:28.670 --> 00:17:35.350
The equation is given by π₯ minus π all squared plus π¦ minus π all squared plus π§ minus π all squared is equal to π squared.
00:17:35.630 --> 00:17:41.860
And given the standard form of the equation of a sphere, we can find both the center of the sphere and the radius of the sphere.
00:17:42.150 --> 00:17:49.180
To find the center of our sphere, the easiest way is to find the values of our variables we would substitute in to make each term equal to zero.
00:17:49.490 --> 00:17:55.090
And to find the radius of this sphere, all we need to do is take the square root of the constant on the right-hand side of the equation.