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Calculus IIChapter 6: 3-Dimensional SpaceSection 6-4

Quadric Surfaces — Question 9

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Question 9

Find and classify the curve of intersection of the sphere x2+y2+z2=25x^2+y^2+z^2=25 with the plane x+2y+2z=9x+2y+2z=9. Find the curve’s center and radius, and determine whether the plane is tangent, secant, or disjoint.

Original worksheet page 1: question and worked solution for 6-4-009
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Question 9 – Solution

Strategy The circle center is the perpendicular projection of the sphere center onto the plane.

See the diagram in the original worksheet below.

Plane distance The origin-to-plane distance is d=9/1+4+4=3d=9/\sqrt{1+4+4}=3. Since 3<53<5, the plane is secant.

Center and radius The foot from the origin lies along ⟨1,2,2⟩\left\langle 1,2,2\right\rangle and is C=99⟨1,2,2⟩=(1,2,2).C=\frac{9}{9}\left\langle 1,2,2\right\rangle=\boxed{(1,2,2)}. By the right triangle from sphere center to circle edge, r2=52−32=16r^2=5^2-3^2=16, so r=4\boxed{r=4}.

Classification The intersection is a circle in the given plane, centered at (1,2,2)(1,2,2) with radius 4.

Original worksheet page 2: question and worked solution for 6-4-009

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

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