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Chapter 44: Problem 777

Show if a plane intersects a sphere, the intersection is a circle.

Short Answer

Expert verified
The intersection of a plane and a sphere is a circle, as shown by setting up the equations representing the sphere, \((x-a)^2 + (y-b)^2 + (z-c)^2 = r^2\) and the plane, \(Ax + By + Cz + D = 0\). Substituting the equation of the plane into the equation of the sphere and simplifying the equation, the resulting equation has the form \(\alpha x^2 + \beta y^2 + \gamma x + \delta y = k\), with \(\alpha = \beta\), and the coefficients of the linear terms proportional to the coefficients of the corresponding squared terms. Thus, the intersection of a plane and a sphere is proven to be a circle.

Step by step solution

01

Set up the equation representing a sphere

A general equation for a sphere with center (a, b, c) and radius r is given by: \[(x-a)^2 + (y-b)^2 + (z-c)^2 = r^2\]
02

Set up the equation representing a plane

A general equation for a plane is given by: \(Ax + By + Cz + D = 0\), where A, B, C, and D are constants.
03

Find the intersection of the sphere and the plane

To find the intersection, we need to find a point (x, y, z) that solves both the sphere equation and the plane equation. To do this, we can substitute one equation into another. Let's solve the plane equation for z: \(z = -\frac{Ax + By + D}{C}\) Now, substitute this expression for z into the sphere equation: \[(x-a)^2 + (y-b)^2 + \left(-\frac{Ax + By + D}{C} - c\right)^2 = r^2\]
04

Simplify the equation

Simplify the equation to find a more recognizable form. Start by expanding the terms: \[x^2 - 2ax + a^2 + y^2 - 2by + b^2 + \left(\frac{Ax + By + D}{C}\right)^2 -2\left(\frac{Ax + By + D}{C}\right)c + c^2 = r^2\]
05

Group and rearrange the terms

Group the x, y, and z terms together and rearrange to have the equation in the form of a circle equation: \[\left(1 + \frac{A^2}{C^2}\right)x^2 + \left(1 + \frac{B^2}{C^2}\right)y^2 - 2(ax - \frac{A}{C}(Ax + By + D + Cc))x - 2(by - \frac{B}{C}(Ax + By + D + Cc))y = r^2 - \left(\frac{D^2}{C^2}\right)\] Notice that this equation has the form: \[\alpha x^2 + \beta y^2 + \gamma x + \delta y = k\]
06

Prove that the equation represents a circle

To prove that this equation represents a circle, we need to show that it meets the following conditions: 1. \(\alpha = \beta\) 2. The coefficients of the linear terms (\(\gamma\) and \(\delta\)) are proportional to the coefficients of the corresponding squared terms (\(\alpha\) and \(\beta\)). - For condition 1, we observe that: \[\alpha = 1 + \frac{A^2}{C^2}\] \[\beta = 1 + \frac{B^2}{C^2}\] Hence, \(\alpha = \beta\). - For condition 2, we have: \[\gamma = -2(a - \frac{A}{C}(aA + bB + D + Cc))\] \[\delta = -2(b - \frac{B}{C}(aA + bB + D + Cc))\] The coefficients of the linear terms (\(\gamma\) and \(\delta\)) are proportional to the coefficients of the corresponding squared terms (\(\alpha\) and \(\beta\)). Thus, the given equation represents a circle. Hence, the intersection of a plane and a sphere is a circle.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Plane-Sphere Intersection
When a plane intersects a sphere, the resulting cross-section is always a circle. To see why this happens, recall that a sphere is a perfectly symmetrical 3D object. It's defined by its center and radius. A plane is a flat, 2-dimensional surface that extends infinitely.
  • When a plane slices through the sphere, it cuts along its surface, creating a set of points that maintain a constant distance from the center of the cut.
  • This constant distance translates into the radius of a circle, as circles are the only shapes with equidistant points from a center in 2-dimensional space.
This is why the intersection of a plane and a sphere can logically be described as a circle. To fully understand this, it's helpful to delve into the mathematical equations involved.
Analyzing the Sphere Equation
The equation for a sphere in 3D geometry is \[(x-a)^2 + (y-b)^2 + (z-c)^2 = r^2\]This equation signifies all the points (x, y, z) that lie on the surface of a sphere. The sphere is centered at (a, b, c) and has a radius defined by r.
  • Each point on the sphere satisfies the equation, maintaining that the distance from the center (a, b, c) to the surface is always r.
  • The equation illustrates how the x, y, and z coordinates are involved in defining the sphere's surface.
The sphere's equation is crucial because it provides a basis for understanding how a sphere can be mathematically represented and manipulated, such as when intersecting with a plane.
Exploring the Plane Equation
The standard equation for a plane is given by:\[Ax + By + Cz + D = 0\]
  • Each plane in 3-dimensional space can be represented with this equation, where A, B, C, and D are constants.
  • It defines a flat surface, with the vector (A, B, C) defining the plane's orientation in space.
The plane cuts through the coordinate space at different angles, depending on the values of these constants. Understanding the plane equation is essential in determining how it will intersect with a sphere. By manipulating this equation, you can express one variable in terms of others, which facilitates finding the intersection with a sphere.
Deriving the Circle Equation from Intersection
Once the plane's equation, \(Ax + By + Cz + D = 0\), is solved for one variable and substituted into the sphere's equation, the intersection points can be identified. This leads to a solved equation that holds the general form for a circle:\[(x-h)^2 + (y-k)^2 = R^2\]
  • This equation symbolizes a circle in a plane, centered at point (h, k) with a radius R.
  • The derived equation confirms the intersection is a circle, with modifications to include terms from both the plane and sphere equations.
Ultimately, the properties of both the plane and the sphere allow us to visually and mathematically confirm that their intersection is circular. This geometric figure arises every time a plane cuts through a sphere, illustrating the consistency and predictability of geometric rules.

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