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Chapter 12: Problem 38

Give a geometric description of the following sets of points. $$x^{2}-4 x+y^{2}+6 y+z^{2}+14=0$$

Short Answer

Expert verified
Answer: The given set of points forms a sphere with center at the point (2, -3, 0) and radius equals 1 unit.

Step by step solution

01

Rewrite the given equation

First, let's rewrite the given equation in a more explicit form and separate the variables: $$x^2 - 4x + y^2 + 6y + z^2 + 14 = 0$$
02

Complete the squares for x, y, and z terms

Complete the square for the x, y, and z terms by adding and subtracting the square of half of their coefficients: $$\left( x^2 - 4x + 2^2 \right) + \left( y^2 + 6y + (-3)^2 \right) + z^2 = -14 + 4 + 9$$ Now, rewrite the equation as follows: $$\left( x - 2 \right)^2 + \left( y + 3 \right)^2 + z^2 = 1$$
03

Identify the geometric shape

The rewritten equation above can now be recognized as the equation of a geometric shape in a 3D coordinate system. The general equation for a sphere in a 3D coordinate system is: $$(x - a)^2 + (y - b)^2 + (z - c)^2 = r^2$$ where (a, b, c) is the center of the sphere, and r is the radius. Comparing our equation with the generic sphere equation, we get: $$\left( x - 2 \right)^2 + \left( y + 3 \right)^2 + z^2 = 1^2$$ Which shows that the given set of points forms a sphere with center (2, -3, 0) and radius 1.
04

Geometric Description

The given set of points forms a sphere with center at the point (2, -3, 0) and radius equals 1 unit.

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