Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 12 - Vectors and the Geometry of Space - Review - Exercises - Page 882: 1


(a) $(x+1)^2+(y-2)^2+(z-1)^2=69$ (b) $(y-2)^2+(z-1)^2=68$ (c) Center: $(4,-1,-3)$ and radius: $ \sqrt {25}=5$

Work Step by Step

(a) $r=\sqrt {(6-(-1))^2+((-2)-2)^2+(3-1)^2}=\sqrt {49+16+4}$ $r=\sqrt {(6-(-1))^2+((-2)-2)^2+(3-1)^2}=\sqrt {69}$ The equation of sphere is: $(x+1)^2+(y-2)^2+(z-1)^2=69$ (b) From part (a): $(x+1)^2+(y-2)^2+(z-1)^2=69$ If it intersects the $yz$-plane , then $x=0$ Thus, $(0+1)^2+(y-2)^2+(z-1)^2=69$ $(y-2)^2+(z-1)^2=69-1$ Hence, $(y-2)^2+(z-1)^2=68$ (c) $x^2+y^2+z^2+-8x+2y+6z+1=0$ $x^2+y^2+z^2+-8x+2y+6z=-1$ Complete the square for $x$, $y$ and $z$. $(x^2-8x+16)+(y^2+2y+1)+(z^2+6z+9)=-1+16+1+9$ $(x-4)^2+(y+1)^2+(z+3)^2=25$ Center: $(4,-1,-3)$ and radius: $ \sqrt {25}=5$
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