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A sphere, Si is given by the equation: r? + y² + 22 – 2x – 4y + 8z + 17 = 0 (a) Write the equation for sphere Si in standard form (b) Sz is a sphere centered at (-3, -2, 1) and tangent to the xy plane. Find the distance from the surface of Si to the surface of S2.

A sphere, Si is given by the equation: r? + y² + 22 – 2x – 4y + 8z + 17 = 0 (a) Write the equation for sphere Si in standard form (b) Sz is a sphere centered at (-3, -2, 1) and tangent to the xy plane. Find the distance from the surface of Si to the surface of S2.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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A sphere, Si is given by the equation: r² + y² + 2² – 2x – 4y +82 + 17 = 0 (a) Write the equation for sphere S, in standard form (b) Sz is a sphere centered at (-3, -2, 1) and tangent to the xy plane. Find the distance from the surface of Si to the surface of S2.
Transcribed Image Text:A sphere, Si is given by the equation: r² + y² + 2² – 2x – 4y +82 + 17 = 0 (a) Write the equation for sphere S, in standard form (b) Sz is a sphere centered at (-3, -2, 1) and tangent to the xy plane. Find the distance from the surface of Si to the surface of S2.
Expert Solution
Step 1

(a)

Consider the given information.

x2+y2+z2-2x-4y+8z+17=0

Now, write the equation in the standard form.

x2-2x+y2-4y+z2+8z+17=0x2-2·x·1+12+y2-2·y·2+22+z2+2·z·4+42=-17+12+22+42x-12+y-22+z+42=-17+1+4+16x-12+y-22+z+42=4

Thus, the standard form of the equation is x-12+y-22+z+42=4

Step 2

(b)

For the sphere S2, the given center is (-3,-2,1). The tangent plane is xy plane in the plane.

So, the coordinate of the plane is x,y,0.

Let the given the point p1 in the plane is (-3,-2,1) and the point p2, which lies in the xy plane, is (-3,-2,0).

Now, calculate the distance between both the points.

P2P1=-3+32+-2+22+1-02=0+0+1=1 

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