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Calculus IIIChapter 1: 3-Dimensional SpaceSection 1-1

The 3-D Coordinate System — Question 3

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Question 3

Consider the fixed points A=(−2,0,0),B=(4,0,0),A=(-2,0,0),\qquad B=(4,0,0), and let P=(x,y,z)P=(x,y,z) satisfy the distance condition d(P,A)=2d(P,B).d(P,A)=2d(P,B).

Tasks

  1. Derive a Cartesian equation for the locus of all possible points PP.

  2. Identify the surface by stating its center and radius.

  3. Determine which of AA and BB lies inside the surface.

  4. Explain geometrically what the factor 22 requires of points on the locus.

Original worksheet page 1: question and worked solution for 1-1-003
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Question 3 – Solution

Strategy Square the distance relation, expand, and complete the square. The resulting surface is an Apollonius sphere.

See the diagram in the original worksheet below.

Derivation The condition is (x+2)2+y2+z2=4((x−4)2+y2+z2).(x+2)^2+y^2+z^2=4\bigl((x-4)^2+y^2+z^2\bigr). After collecting terms and dividing by 33, x2−12x+y2+z2+20=0.x^2-12x+y^2+z^2+20=0. Completing the square gives (x−6)2+y2+z2=16.\boxed{(x-6)^2+y^2+z^2=16}. Thus the locus is the sphere centered at (6,0,0)(6,0,0) with radius 44.

Interpretation Point BB is 22 units from the center and lies inside; AA is 88 units from the center and lies outside. A point must remain closer to BB because its distance to AA is required to be twice as large.

Verification At the axial endpoints (2,0,0)(2,0,0) and (10,0,0)(10,0,0), the distance pairs are (4,2)(4,2) and (12,6)(12,6), respectively, confirming the ratio.

Original worksheet page 2: question and worked solution for 1-1-003

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