Video Transcript
Find the equation of a sphere that passes through the points 𝐴 zero, three, negative two and 𝐵 negative one, negative three, negative five given that its center lies on the 𝑧-axis.
We recall that the Cartesian equation of a sphere with radius 𝑟 and center 𝑎𝑏𝑐 written in standard form is 𝑥 minus 𝑎 all squared plus 𝑦 minus 𝑏 all squared plus 𝑧 minus 𝑐 all squared is equal to 𝑟 squared. In this question, we are told that the center of the sphere lies on the 𝑧-axis. This means that both the 𝑥- and 𝑦-coordinates of the center will be zero. Substituting this into the standard form, we have 𝑥 minus zero all squared plus 𝑦 minus zero all squared plus 𝑧 minus 𝑐 all squared is equal to 𝑟 squared. This simplifies to 𝑥 squared plus 𝑦 squared plus 𝑧 minus 𝑐 all squared is equal to 𝑟 squared.
We are also told that our sphere passes through the points 𝐴 and 𝐵. This means that we can substitute the coordinates of these points into our equation. At point 𝐴, we have 𝑥 is equal to zero, 𝑦 is equal to three, and 𝑧 is equal to negative two. Substituting these values gives us the following equation. In order to square negative two minus 𝑐, we multiply negative two minus 𝑐 by negative two minus 𝑐. We can then distribute the parentheses or expand the brackets using the FOIL method. This gives us four plus two 𝑐 plus two 𝑐 plus 𝑐 squared. And collecting like terms, this is equal to four plus four 𝑐 plus 𝑐 squared. As three squared is equal to nine, our equation becomes nine plus four plus four 𝑐 plus 𝑐 squared is equal to 𝑟 squared. As nine plus four is 13, we have 13 plus four 𝑐 plus 𝑐 squared is equal to 𝑟 squared. As there are two unknowns in this equation, we will call this equation one and now consider point 𝐵.
This point has coordinates negative one, negative three, negative five. Substituting these into our equation, we have negative one squared plus negative three squared plus negative five minus 𝑐 all squared is equal to 𝑟 squared. Squaring negative one gives us one and squaring negative three gives us nine. Squaring negative five minus 𝑐 in the same way as we squared negative two minus 𝑐 gives us 25 plus 10𝑐 plus 𝑐 squared. This is all equal to 𝑟 squared. Simplifying our equation, we have 35 plus 10𝑐 plus 𝑐 squared is equal to 𝑟 squared. We will call this equation two.
Since the right-hand side of our equations are equal, the left-hand sides must also be equal. This means that 13 plus four 𝑐 plus 𝑐 squared must be equal to 35 plus 10𝑐 plus 𝑐 squared. We can subtract 𝑐 squared from both sides. We can then subtract four 𝑐 and 35 from both sides of the equation, giving us negative 22 is equal to six 𝑐. And dividing through by six, 𝑐 is equal to negative 22 over six, which simplifies to negative 11 over three.
We now have the left-hand side of the equation of our sphere. It is equal to 𝑥 squared plus 𝑦 squared plus 𝑧 plus 11 over three all squared. We can now substitute 𝑐 is equal to negative 11 over three into equation one or equation two to calculate 𝑟 squared. If we use equation one, we have 13 plus four multiplied by negative 11 over three plus negative 11 over three squared is equal to 𝑟 squared. The left-hand side simplifies to 13 minus 44 over three plus 121 over nine. Creating a common denominator of nine, we have 𝑟 squared is equal to 117 over nine minus 132 over nine plus 121 over nine. This is equal to 106 over nine.
We now have the equation of a sphere that passes through the points zero, three, negative two and negative one, negative three, negative five with a center that lies on the 𝑧-axis. It is 𝑥 squared plus 𝑦 squared plus 𝑧 plus 11 over three all squared is equal to 106 over nine.
