# spherical

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• September 5th 2008, 06:26 AM
Apprentice123
spherical
Doubts with the response

To determine the equation of the sphere that passes in A and B, and has straight center in s
$A(0,4,3)$ $B(1,1,-5)$
(s):
$x=z+2$
$y=z-3$

My solution:
For the straight, center: $C(1,1,1)$

$R=d(C,A)=\sqrt{14}$

The equation is:
$(x-1)^2+(y-1)^2+(z-1)^2=14$

BUT THE ANSWER IS:
$(x-3)^2+(y+2)^2+(z-1)^2=49$
• September 5th 2008, 07:04 AM
Laurent
How did you find the center ? It is supposed to lie on the line S, so it must satisfy both equations, which it does'nt.

The center must be equidistant from A and B, and satisfy the two equations. This makes 3 equations, allowing you to find the three components of the center.

Laurent.
• September 5th 2008, 09:20 AM
Plato
Each of the points is the same distance from the center.
So solve $\left( {z + 2} \right)^2 + \left( {z - 7} \right)^2 + \left( {z - 8} \right)^2 = \left( {z + 1} \right)^2 + \left( {z - 4} \right)^2 + \left( {z + 5} \right)^2$.
If correct, you will find z=1.
Then you can find x&y on s.
• September 5th 2008, 09:28 AM
Laurent
How do you get this equation, Plato ?

A and B are at the same distance from the center O, so the coordinates $(x,y,z)$ of the center satisfy:

$x^2+(y-4)^2+(z-3)^2=(x-1)^2+(y-1)^2+(z+5)^2.$

Expand this. The squares simplify, so you get a linear equation. The equations of S provide two others, and you have a linear system of three equations in three unknowns.
• September 5th 2008, 09:31 AM
Plato
Quote:

Originally Posted by Laurent
How do you get this equation, Plato ?

It is easy to see. The center is on s. On s, x & y are defined in terms of z.
• September 5th 2008, 09:42 AM
Laurent
Well done, Plato ! This is indeed quicker to write down this way.
• September 6th 2008, 10:29 AM
Apprentice123
thanks