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Chapter 26: Problem 137

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 [2003] (a) 4 (b) 1 (c) 2 (d) 3

Short Answer

Expert verified
The radius of the circle is 4, option (a).

Step by step solution

01

Express the Equation of the Sphere in Center-Radius Form

The given equation of the sphere is \(x^2 + y^2 + z^2 + 2x - 2y - 4z - 19 = 0\). First, we need to complete the square for each variable to express the equation in the center-radius form: \((x-a)^2 + (y-b)^2 + (z-c)^2 = R^2\).
02

Completing the Square

Rewrite the terms individually:\[x^2 + 2x = (x+1)^2 - 1,\]\[y^2 - 2y = (y-1)^2 - 1,\]\[z^2 - 4z = (z-2)^2 - 4.\]Incorporate these into the original equation,\((x+1)^2 - 1 + (y-1)^2 - 1 + (z-2)^2 - 4 - 19 = 0,\)which simplifies to \((x+1)^2 + (y-1)^2 + (z-2)^2 = 25.\)
03

Identify the Sphere's Center and Radius

The sphere equation \((x+1)^2 + (y-1)^2 + (z-2)^2 = 25\) has a center \((-1, 1, 2)\) and a radius of 5 (since \(R^2=25\)).
04

Determine the Distance from the Center to the Plane

The plane equation is \(x + 2y + 2z + 7 = 0\). Use the point-to-plane distance formula:\[d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}},\]where \((a, b, c) = (1, 2, 2)\) and \((x_1, y_1, z_1) = (-1, 1, 2)\). Calculate:\[d = \frac{|1(-1) + 2(1) + 2(2) + 7|}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{|9|}{\sqrt{9}} = 3.\]
05

Calculate the Radius of the Circle

The plane cuts a circle in the sphere, and this circle's radius is found using the relationship: \(r = \sqrt{R^2 - d^2}\). With \(R = 5\) and \(d = 3\), we find:\[r = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4.\]
06

Conclude with the Result

The radius of the circle formed by the intersection of the sphere and plane is \(4\). Therefore, the answer is (a) 4.

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Key Concepts

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

Sphere Equation
The sphere equation is a mathematical representation in three-dimensional space that describes a set of points at a constant distance, called the radius, from a fixed point, known as the center. The general equation of a sphere in three dimensions is given by:
  • \(x^2 + y^2 + z^2 + 2x - 2y - 4z - 19 = 0\)
This equation is expanded and needs simplification to determine the center and radius of the sphere. To solve this, one effective method used is 'completing the square'. By doing so, we can transform it into a more recognizable form.
Center-Radius Form
The center-radius form of a sphere's equation makes it easy to visualize the geometric characteristics of the sphere. It is expressed as:
  • \((x-a)^2 + (y-b)^2 + (z-c)^2 = R^2\)
Where
  • \((a, b, c)\) is the center of the sphere.
  • \(R\) is the radius.
To convert a given quadratic form into the center-radius form, we complete the square for each variable. This process helps identify the center by isolating the terms pertaining to each coordinate, and the constant on the right side gives the square of the radius.
Point-to-Plane Distance
The point-to-plane distance is a way to compute how far a point in space is from a given plane. To find this distance, we can use the formula:
  • \[d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}\]
In this formula:
  • \((a, b, c)\) are the coefficients of the plane equation.
  • \((x_1, y_1, z_1)\) are the coordinates of the point.
  • \(d\) is the constant in the plane equation.
This metric helps us determine whether the intersection of the sphere and the plane results in an empty set, a point, or a circle. A distance less than the sphere's radius indicates that a circle is formed.
Completing the Square
Completing the square is a powerful algebraic technique used to rearrange a quadratic equation into an easily interpretable form. For a sphere's equation, it involves isolating each squared term to identify the center and calculate the radius. Consider the expressions:
  • \(x^2 + 2x,\)
  • \(y^2 - 2y,\)
  • \(z^2 - 4z\)
Each of these can be rewritten as perfect squares:
  • \((x+1)^2 - 1\)
  • \((y-1)^2 - 1\)
  • \((z-2)^2 - 4\)
After combining and simplification, they help in deriving the center-radius form. This transformation simplifies multiple algebraic processes, making the problem more approachable and solvable.

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Most popular questions from this chapter

The coordinates of the foot of the perpendicular from the point \((1,-2,1)\) on the plane containing the lines, \(\frac{x+1}{6}=\frac{y-1}{7}=\frac{z-3}{8}\) and \(\frac{x-1}{3}=\frac{y-2}{5}=\frac{z-3}{7}\), is: \(\quad\) Online April 8, 2017] (a) \((2,-4,2)\) (b) \((-1,2,-1)\) (c) \((0,0,0)\) (d) \((1,1,1)\)

A perpendicular is drawn from a point on the line \(\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z}{1}\) to the plane \(x+y+z=3\) such that the foot of the perpendicular \(\mathrm{Q}\) also lies on the plane \(x-y+z=3\). Then the co-ordinates of Q are : [April 10, 2019 (II)](a) \((1,0,2)\) (b) \((2,0,1)\) (c) \((-1,0,4)\) (d) \((4,0,-1)\)

If \(L_{1}\) is the line of intersection of the planes \(2 x-2 y+3 z-2=0, x-y+z+1=0\) and \(L_{2}\) is the line of intersection of the planes \(x+2 y-z-3=0\), \(3 x-y+2 z-1=0\), then the distance of the origin from the plane, containing the lines \(\mathrm{L}_{1}\) and \(\mathrm{L}_{2}\), is : [2018] (a) \(\frac{1}{3 \sqrt{2}}\) (b) \(\frac{1}{2 \sqrt{2}}\) (c) \(\frac{1}{\sqrt{2}}\) (d) \(\frac{1}{4 \sqrt{2}}\)

The shortest distance between the lines \(\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1}\) and \(x+y+z+1=0,2 x-y+z+3=0\) is: \([\operatorname{Sep} .06,2020(I)]\) (a) 1 (b) \(\frac{1}{\sqrt{3}}\) (c) \(\frac{1}{\sqrt{2}}\) (d) \(\frac{1}{2}\)

If the lines \(\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}\) and \(\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}\) are coplanar, then \(k\) can have [2013] (a) any value (b) exactly one value (c) exactly two values (d) exactly three values
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