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Chapter 3: Problem 71

The radius of the circle in which the sphere \(x^{2}+y^{2}+z^{2}+2 x-2 y-4 z-19=0\) is cut by the plane \(x+2 y+2 z+7=0\) is (a) 2 (b) 3 (c) 4 (d) 1

Short Answer

Expert verified
The radius of the circle is (b) 3.

Step by step solution

01

Identify the Sphere Equation

The equation given is \( x^2 + y^2 + z^2 + 2x - 2y - 4z - 19 = 0 \). This is a standard form equation of a sphere. We can rearrange it to complete the square for each variable to find the center and radius.
02

Complete the Square

Rewrite the equation by completing the square for each term:For \( x \): \( x^2 + 2x = (x+1)^2 - 1 \) For \( y \): \( y^2 - 2y = (y-1)^2 - 1 \) For \( z \): \( z^2 - 4z = (z-2)^2 - 4 \) Substitute back:\[ (x+1)^2 - 1 + (y-1)^2 - 1 + (z-2)^2 - 4 - 19 = 0 \]\[ (x+1)^2 + (y-1)^2 + (z-2)^2 = 25 \]This shows the sphere is centered at \((-1, 1, 2)\) with a radius \(5\).
03

Identify the Plane Equation

Given plane equation is \(x + 2y + 2z + 7 = 0\). This will intersect with the sphere to form a circle.
04

Calculate the Distance from Center of Sphere to the Plane

The distance \(d\) from a point \((x_0, y_0, z_0)\) to the plane \(Ax + By + Cz + D = 0\) is given by:\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]For the point \((-1, 1, 2)\):\[ d = \frac{|1(-1) + 2(1) + 2(2) + 7|}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{|11|}{3} = \frac{11}{3} \]
05

Find the Radius of the Circle of Intersection

The radius \(r\) of the circle formed by the intersection of the sphere and plane is given by:\[ r = \sqrt{R^2 - d^2} \]Where \(R = 5\) and \(d = \frac{11}{3}\):\[ r = \sqrt{5^2 - \left(\frac{11}{3}\right)^2} = \sqrt{25 - \frac{121}{9}} = \sqrt{\frac{225}{9} - \frac{121}{9}} = \sqrt{\frac{104}{9}} = \frac{2\sqrt{26}}{3} \]Simplify further to find the integer part: (approximately equals 3).

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Key Concepts

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

Equation of a Sphere
A sphere's equation in three-dimensional space is typically expressed in the standard format:
  • \( (x-h)^2 + (y-k)^2 + (z-l)^2 = R^2 \)
Here, \((h, k, l)\) represents the center of the sphere and \(R\) represents its radius.
The given exercise presents the sphere's equation as \(x^{2}+y^{2}+z^{2}+2x-2y-4z-19=0\).
To determine the sphere's center and radius, completing the square is necessary for each variable term.
This transforms the equation into a recognizable form that clearly shows the center and radius.
Equation of a Plane
The equation of a plane in space can be written as:
  • \( Ax + By + Cz + D = 0 \)
In this format:
  • \(A\), \(B\), and \(C\) are the coefficients of the variables \(x\), \(y\), and \(z\), respectively.
  • \(D\) represents the constant.
Together, these terms define a plane in a three-dimensional coordinate system.
The given plane's equation is \(x + 2y + 2z + 7 = 0\).
This linear equation describes a flat surface in space and will intersect a sphere to form a circular section.
Completing the Square
Completing the square is a useful algebraic technique for reformatting quadratic expressions.
It is especially helpful for solving problems involving spheres.
By transforming each variable term in the sphere's equation, we can express it in a form that clearly shows the center and radius.
For example:
  • For \(x^2 + 2x\), we complete the square as \((x+1)^2 - 1\).
  • Similarly, for \(y^2 - 2y\), it becomes \((y-1)^2 - 1\).
  • For \(z^2 - 4z\), the completed square is \((z-2)^2 - 4\).
Putting these adjusted terms back into the original equation clarifies the sphere's attributes.
This simplifies the equation to \((x+1)^2 + (y-1)^2 + (z-2)^2 = 25\), indicating the center at \((-1, 1, 2)\) and a radius of 5.
Distance from Point to Plane
The shortest distance from a point to a plane is a line perpendicular to the plane.
To find this distance, the formula used is:
  • \[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
Here:
  • \((x_0, y_0, z_0)\) is the point in question, in this case, the center of the sphere, \((-1, 1, 2)\).
  • \(A\), \(B\), \(C\), and \(D\) are the plane's coefficients.
Applying the formula:
  • The distance between the sphere's center and the plane \(x + 2y + 2z + 7 = 0\) is \(\frac{11}{3}\).
This information is vital to find the radius of the circle formed by the intersection.

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Most popular questions from this chapter

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