Radius = r

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Figure 17-4 A sphere of radius r in Cartesian xyz space, centered

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away from the origin All points on the sphere s surface are at distance r from the center point (x0,y0,z0)

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Surfaces in Three-Space

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Squaring both sides of this equation and then transposing the left- and right-hand sides, we obtain (xp x0)2 + (yp y0)2 + (zp z0)2 = r2 Every point on the sphere s surface is the same distance from P as (x0,y0,z0), so we can generalize to get (x x0)2 + (y y0)2 + (z z0)2 = r2 This is the standard-form equation for a sphere of radius r, centered at the point (x0,y0,z0) in Cartesian xyz space

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An example Suppose we have a sphere whose center is at the origin, and whose radius is 7 units If we let r = 7 in the general equation for a sphere centered at the origin, then we have

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x2 + y2 + z2 = 72 which can be simplified to x2 + y2 + z2 = 49

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Another example Consider a sphere centered at the point ( 2,4, 1) with a radius of 5 units in Cartesian xyz space We can let

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x0 = 2 y0 = 4 z0 = 1 r=5 in the general equation for a sphere centered at a point other than the origin When we plug in the numbers, we obtain [x ( 2)]2 + (y 4)2 + [z ( 1)]2 = 52 which simplifies to (x + 2)2 + (y 4)2 + (z + 1)2 = 25

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Spheres

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Are you confused

The radius of a sphere is usually defined as a positive real number If we define the radius of a particular sphere as a negative real number, we get the same equation as we would if we defined the radius as the absolute value of that number That s because we square the radius when we work out the formula For example, if we have a sphere centered at the origin with a radius of 4 units, then its equation is x2 + y2 + z2 = 42 which simplifies to x2 + y2 + z2 = 16 If we find a companion antisphere centered at the origin with radius 4 units, then its equation is x2 + y2 + z2 = ( 4)2 which also simplifies to x2 + y2 + z2 = 16 In physics and engineering, it s possible to come up with spheres having negative radii, as well as negative dimensions for other physical objects These results are usually mere artifacts of the calculation process, and don t have any significance in the real world However, if you ever encounter a sphere whose radius is represented by an imaginary number such as j4, then you have good reason to be confused until you know what sort of object or phenomenon the equation describes!

Here s a challenge!

Suppose we re told that the following equation represents a sphere in Cartesian xyz space: x2 6x + y2 2y + z2 + 4z = 86 We re also informed that the sphere has a radius of 10 units What are the coordinates of the center of the sphere

Solution

To solve this problem, we need some intuition We know that the radius r of the sphere is 10 units Therefore, r2 = 100 If we add 14 to both sides of the original equation, we get r2 on the right-hand side: x2 6x + y2 2y + z2 + 4z + 14 = 100 We can split the stand-alone constant 14 into the sum of 9, 1, and 4, getting x2 6x + y2 2y + z2 + 4z + 9 + 1 + 4 = 100

Surfaces in Three-Space

Rearranging the addends on the left-hand side produces the following equation: x2 6x + 9 + y2 2y + 1 + z2 + 4z + 4 = 100 When we group the terms on the left-hand side by threes, we obtain (x2 6x + 9) + (y2 2y + 1) + (z2 + 4z + 4) = 100 This is a sum of three perfect squares! Factoring them individually gives us (x 3)2 + (y 1)2 + (z + 2)2 = 100 The coordinates of the center point are therefore x0 = 3 y0 = 1 z0 = 2 Expressed as an ordered triple, it s (3,1, 2)