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Calculus IIIChapter 1: 3-Dimensional SpaceSection 1-1

The 3-D Coordinate System — Question 7

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Question 7

Find every sphere satisfying all of the following conditions:

  1. The sphere lies entirely in the octant x≥0,y≤0,z≥0.x\ge 0,\qquad y\le 0,\qquad z\ge 0.

  2. It is tangent to each of the three coordinate planes.

  3. It passes through the point P=(1,−2,3).P=(1,-2,3).

For each sphere, state its center, radius, and Cartesian equation. Then give a geometric explanation for why two different spheres satisfy the same conditions.

Original worksheet page 1: question and worked solution for 1-1-007
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Question 7 – Solution

Strategy Tangency to all three coordinate planes makes the center’s coordinate magnitudes equal to the radius. The specified octant fixes their signs.

See the diagram in the original worksheet below.

Center condition If the radius is r>0r>0, then the center must be C=(r,−r,r).C=(r,-r,r). Since PP lies on the sphere, (1−r)2+(−2+r)2+(3−r)2=r2.(1-r)^2+(-2+r)^2+(3-r)^2=r^2. Expanding gives 2r2−12r+14=02r^2-12r+14=0, or r2−6r+7=0.r^2-6r+7=0. Hence r=3±2r=3\pm\sqrt 2, both positive.

Results The two spheres are (x−r)2+(y+r)2+(z−r)2=r2,r=3±2.\boxed{(x-r)^2+(y+r)^2+(z-r)^2=r^2,\qquad r=3\pm\sqrt 2}.

Why two? Along the line of possible centers (r,−r,r)(r,-r,r), the distance to PP first decreases and then increases, so the equation “distance to center equals rr” can be met at two scales. Direct substitution verifies both roots, and each sphere has extreme coordinates touching—but not crossing—the three coordinate planes.

Original worksheet page 2: question and worked solution for 1-1-007

Original worksheet layout. Use Enlarge or open the PDF for a closer view.

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