Help - Equation of a sphere in 3D

**rain21** · January 9th 2010, 07:31 PM

the problem:
Find the equation of the sphere that contains the points A = (2,4,4) , B= (4,2,0) , and C = (4,0,2), if its center lies on the plane y=x.

My attempt:

|AB|= sqroot(24)
|BC|= sqroot(8)
|AC|= sqroot(24)

from this i guessed that the center of the circle was point A, since points B and C were both equal distances away from them.
which gives me an r = sqroot(24)

Thus, the equation of the sphere is:

(x-2)^2 + (y-4)^2 + (z-4)^2 = 24

My problem is... the center (2,4,4) is not on the plane y=x

I'm confused. any help?

**Black** · January 9th 2010, 08:30 PM

Let $(x_0, x_0, z_0)$ be the center of the sphere. Choosing point $B$ , the radius of the sphere is

$r=\sqrt{(x_0-4)^2+(x_0-2)^2+z_0^2}$ .

Then the equation of the sphere is

$(x-x_0)^2+(y-x_0)^2+(z-z_0)^2=(x_0-4)^2+(x_0-2)^2+z_0^2$ .

Plug in points $A$ and $C$ to get the system of equations

$(2-x_0)^2+(4-x_0)^2+(4-z_0)^2=(x_0-4)^2+(x_0-2)^2+z_0^2$
$(4-x_0)^2+x_0^2+(2-z_0)^2=(x_0-4)^2+(x_0-2)^2+z_0^2$ .

Then just solve for $(x_0,z_0)$ and put the values into the equation of the sphere to get your answer.

**rain21** · January 9th 2010, 09:05 PM

Originally Posted by Black

Let $(x_0, x_0, z_0)$ be the center of the sphere. Choosing point $B$ , the radius of the sphere is

$r=\sqrt{(x_0-4)^2+(x_0-2)^2+z_0^2}$ .

Then the equation of the sphere is

$(x-x_0)^2+(y-x_0)^2+(z-z_0)^2=(x_0-4)^2+(x_0-2)^2+z_0^2$ .

Plug in points $A$ and $C$ to get the system of equations

$(2-x_0)^2+(4-x_0)^2+(4-z_0)^2=(x_0-4)^2+(x_0-2)^2+z_0^2$
$(4-x_0)^2+x_0^2+(2-z_0)^2=(x_0-4)^2+(x_0-2)^2+z_0^2$ .

Then just solve for $(x_0,z_0)$ and put the values into the equation of the sphere to get your answer.

thank you.

was point B chosen to find the radius because there was no value for z?

**Black** · January 9th 2010, 09:16 PM

Yeah, I chose B because of the of the 0 in the z-coordinate. Choosing A or C to find the radius will still lead you to the same answer.

**HallsofIvy** · January 10th 2010, 05:39 AM

Originally Posted by rain21

the problem:
Find the equation of the sphere that contains the points A = (2,4,4) , B= (4,2,0) , and C = (4,0,2), if its center lies on the plane y=x.

My attempt:

|AB|= sqroot(24)
|BC|= sqroot(8)
|AC|= sqroot(24)

from this i guessed that the center of the circle was point A, since points B and C were both equal distances away from them.
which gives me an r = sqroot(24)

Well, there is your problem! You are told that the point A is on the sphere. It can't be the center of the sphere! Also, obviously, A does NOT "lie on the plane y=x". If you must assume things, don't assume things that contradict the information you are given!

You need to find the center. Do you remember how to find the center of a circle given three points on the circle? The perpendicular bisector of each segment through to points must intersect at the center. For three dimensions, the corresponding idea is that the planes orthogonal to the line segments. The segment from A = (2,4,4) to B= (4,2,0) has midpoint (3, 3, 2) and "direction vector" $2\vec{i}- 4\vec{k}$ so the plane orthogonal to that segment and containing (3, 3, 2) is 2(x- 3)- 4(z- 2)= 0. The segment from A= (2, 4, 4) to C = (4,0,2) has midpoint (3, 2, 3) and direction vector $2\vec{i}- 4\vec{j}- 2\vec{k}$ so the plane orthogonal to the segment and containing (3, 2, 3) is 2(x- 3)- 4(y- 2)- 2(z- 3)= 0. That is not enough to determine the center of the sphere (using the segment from B to C will not give you anything new) which is why they told you "its center lies on the plane y=x".

Solve the three equations 2(x- 3)- 4(z- 2)= 0, 2(x- 3)- 4(y- 2)- 2(z- 3)= 0, and y= x for (x, y, z), the center of the sphere.

Thus, the equation of the sphere is:

(x-2)^2 + (y-4)^2 + (z-4)^2 = 24

My problem is... the center (2,4,4) is not on the plane y=x

I'm confused. any help?

**rain21** · January 10th 2010, 06:52 PM

Originally Posted by Black

Yeah, I chose B because of the of the 0 in the z-coordinate. Choosing A or C to find the radius will still lead you to the same answer.

I followed everything you advised, but wouldn't the radius be: $r=\sqrt{(4-x_0)^2+(2-x_0)^2+z_0^2}$ .

I got a center of (2,2,2) and a radius = $\sqrt{8}$

=/

**Black** · January 11th 2010, 07:26 AM

Originally Posted by rain21

I got a center of (2,2,2) and a radius = $\sqrt{8}$

=/

Yeah, that's it.

You can always check your answer by plugging in the three points into the equation.

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