Answer
(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 the 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$
