# Math Help - Finding the equation of a circle given it pases through the 4 vertices of a pyramid?

1. ## Finding the equation of a circle given it pases through the 4 vertices of a pyramid?

In 3D Euclidian Vector geometry, if given the 4 vertices of a (Triangular) pyramid:

A (12,0,0) B(0,6,0) C(0,0,4) and O (origin: 0,0,0)

how can you find the sphere that passes through those vertices?

NB: this sphere is S: x² + y² + z² -12x -6y -4z = 0 but i want to know HOW you find this??

Help VERY appreciated thanks!

2. Originally Posted by Yehia
In 3D Euclidian Vector geometry, if given the 4 vertices of a (Triangular) pyramid:

A (12,0,0) B(0,6,0) C(0,0,4) and O (origin: 0,0,0)

how can you find the sphere that passes through those vertices?

NB: this sphere is S: x² + y² + z² -12x -6y -4z = 0 but i want to know HOW you find this??
let $(a,b,c)$ be the sphere's center of radius $r$. each vertex is equidistant from the center and yields the following four equations ...

$r^2 = (a-12)^2 + b^2 + c^2$

$r^2 = a^2 + (b-6)^2 + c^2$

$r^2 = a^2 + b^2 + (c-4)^2$

$r^2 = a^2 + b^2 + c^2$

using each of the first three equations and the last one, you get these equations ...

$144 - 24a = 0$ ... $a = 6$

$36 - 12b = 0$ ... $b = 3$

$16 - 8c = 0$ ... $c = 2$

sphere's equation is

$(x - 6)^2 + (y - 3)^2 + (z - 2)^2 = 49$

expand and simplify to get the equation you provided

3. Originally Posted by skeeter
let $(a,b,c)$ be the sphere's center of radius $r$. each vertex is equidistant from the center and yields the following four equations ...

$r^2 = (a-12)^2 + b^2 + c^2$

$r^2 = a^2 + (b-6)^2 + c^2$

$r^2 = a^2 + b^2 + (c-4)^2$

$r^2 = a^2 + b^2 + c^2$

using each of the first three equations and the last one, you get these equations ...

$144 - 24a = 0$ ... $a = 6$

$36 - 12b = 0$ ... $b = 3$

$16 - 8c = 0$ ... $c = 2$

sphere's equation is

$(x - 6)^2 + (y - 3)^2 + (z - 2)^2 = 49$

expand and simplify to get the equation you provided
Sorry, but how did you get those four equations? (Probably me just not seeing the wood for the trees).

Thanks for the help.

4. Originally Posted by Yehia
Sorry, but how did you get those four equations? (Probably me just not seeing the wood for the trees).

Thanks for the help.
distance formula between any two points in space ...

$r = \sqrt{(x-a)^2 + (y-b)^2 + (z-c)^2}$

5. Hello, Yehia!

I'll give it a try . . .

In 3-space, we are given the four vertices of a tetrahedron:
. . . . $A(12,0,0),\;B(0,6,0),\;C(0,0,4),\;O(0,0,0)$

Find the sphere that passes through those vertices

Answer: . $x^2 + y^2 + z^2 -12x -6y -4z \:=\:0$

The center $\,P$ is equidistant from all four points: . $PA \,=\, PB \,=\,PC\,=\,PD$

$\underbrace{\sqrt{(x-12)^2+y^2}}_{[1]} \:=\:\underbrace{\sqrt{x^2+(y-6)^2 + z^2}}_{[2]}$

. . . . . . . . . . . . . . $=\:\underbrace{\sqrt{x^2 + y^2 + (z-4)^2}}_{[3]} \:=\:\underbrace{\sqrt{x^2+y^2+z^2}}_{[4]}$

$\text{Set }[1] = [4]\!:\;x^2 - 24x + 144 + y^2 + z^2 \:=\:z^2+y^2+z^2$

. . . . . . . . . $-24x + 144 \:=\:0 \quad\Rightarrow\quad\boxed{ x \:=\:6}$

$\text{Set }[2] = [4]\!:\;x^2 + y^2 - 12y + 36 + z^2 \:=\:x^2 + y^2+z^2$

. . . . . . . . . $-12y + 36 \:=\:0 \quad\Rightarrow\quad\boxed{ y \:=\:3}$

$\text{Set }[3] = [4]\!:\;\;x^2+y^2+z^2-8z + 16 \:=\:x^2+y^2+z^2$

. . . . . . . . . $-8z+16 \:=\:0 \quad\Rightarrow\quad\boxed{ z \:=\:2}$

Hence, the center is: $P(6,3,2)$

The distance from $\,P$ to any vertex is the radius.

. . $r \:=\:PO \:=\:\sqrt{(6\!-\!0)^2 + (3\!-\!0)^2 + (2\!-\!0)^2} \:=\:\sqrt{36 + 9 + 4} \:=\:\sqrt{49} \;=\;7$

The equation of the sphere is: . $(x-6)^2 + (y-3)^2 + (z-2)^2 \:=\:7^2$

6. Skeeter and Soroban are using the fact that the distance from the center of the sphere to each of the given points is the same. Here's another way to argue that gives essentially the same equations: the general equation of a sphere is $(x- a)^2+ (y- b)^2+ (z- c)^2= r^2$.

To have (12,0,0) on the sphere, we must have
$(12- a)^2+ (0- b)^2+ (0- c)^2= r^2$.

To have (0,6,0) on the sphere, we must have
$(0- a)^2+ (6- b)^2+ c^2= r^2$.

To have (0,0,4) on the sphere, we must have
$(0- a)^2+ (0- b)^2+ (4- c)^2= r^2$.

To have (0,0,0) on the sphere, we must have
$(0- a)^2+ (0- b)^2+ (0- c)^2= r^2$.

That gives four equations to solve for the four values, a, b, c, and r.

When you multiply out the first three equations, each will have $a^2+ b^2+ c^2$ plus linear terms equal to $r^2$. If you replace $a^2+ b^2+ c^2$ by $r^2$ from the fourth equation, each equation immediately reduces to a simple equation.

7. Originally Posted by skeeter
distance formula between any two points in space ...

$r = \sqrt{(x-a)^2 + (y-b)^2 + (z-c)^2}$
Originally Posted by Soroban
Hello, Yehia!

I'll give it a try . . .

The center $\,P$ is equidistant from all four points: . $PA \,=\, PB \,=\,PC\,=\,PD$

$\underbrace{\sqrt{(x-12)^2+y^2}}_{[1]} \:=\:\underbrace{\sqrt{x^2+(y-6)^2 + z^2}}_{[2]}$

. . . . . . . . . . . . . . $=\:\underbrace{\sqrt{x^2 + y^2 + (z-4)^2}}_{[3]} \:=\:\underbrace{\sqrt{x^2+y^2+z^2}}_{[4]}$

$\text{Set }[1] = [4]\!:\;x^2 - 24x + 144 + y^2 + z^2 \:=\:z^2+y^2+z^2$

. . . . . . . . . $-24x + 144 \:=\:0 \quad\Rightarrow\quad\boxed{ x \:=\:6}$

$\text{Set }[2] = [4]\!:\;x^2 + y^2 - 12y + 36 + z^2 \:=\:x^2 + y^2+z^2$

. . . . . . . . . $-12y + 36 \:=\:0 \quad\Rightarrow\quad\boxed{ y \:=\:3}$

$\text{Set }[3] = [4]\!:\;\;x^2+y^2+z^2-8z + 16 \:=\:x^2+y^2+z^2$

. . . . . . . . . $-8z+16 \:=\:0 \quad\Rightarrow\quad\boxed{ z \:=\:2}$

Hence, the center is: $P(6,3,2)$

The distance from $\,P$ to any vertex is the radius.

. . $r \:=\:PO \:=\:\sqrt{(6\!-\!0)^2 + (3\!-\!0)^2 + (2\!-\!0)^2} \:=\:\sqrt{36 + 9 + 4} \:=\:\sqrt{49} \;=\;7$

The equation of the sphere is: . $(x-6)^2 + (y-3)^2 + (z-2)^2 \:=\:7^2$

Originally Posted by HallsofIvy
Skeeter and Soroban are using the fact that the distance from the center of the sphere to each of the given points is the same. Here's another way to argue that gives essentially the same equations: the general equation of a sphere is $(x- a)^2+ (y- b)^2+ (z- c)^2= r^2$.

To have (12,0,0) on the sphere, we must have
$(12- a)^2+ (0- b)^2+ (0- c)^2= r^2$.

To have (0,6,0) on the sphere, we must have
$(0- a)^2+ (6- b)^2+ c^2= r^2$.

To have (0,0,4) on the sphere, we must have
$(0- a)^2+ (0- b)^2+ (4- c)^2= r^2$.

To have (0,0,0) on the sphere, we must have
$(0- a)^2+ (0- b)^2+ (0- c)^2= r^2$.

That gives four equations to solve for the four values, a, b, c, and r.

When you multiply out the first three equations, each will have $a^2+ b^2+ c^2$ plus linear terms equal to $r^2$. If you replace $a^2+ b^2+ c^2$ by $r^2$ from the fourth equation, each equation immediately reduces to a simple equation.
Thank you guys very very much, yes I do understand it now, like I said, i couldn't see the wood for the trees. But if you don't mind I just have 3 other questions (still reffering to this question though).

How do you show that the centre of this sphere lies OUTSIDE the pyramid??
and also how do you prove that a point lies inside a sphere?
and one last thing: how would you find the point on a sphere that lies CLOSEST to a point P that is inside that sphere.

I hope those that sounds clear and again thanks very much!

8. Originally Posted by Yehia
In 3D Euclidean Vector geometry, if given the 4 vertices of a (Triangular) pyramid:

A (12,0,0) B(0,6,0) C(0,0,4) and O (origin: 0,0,0)

how can you find the sphere that passes through those vertices?

NB: this sphere is S: x² + y² + z² -12x -6y -4z = 0 but i want to know HOW you find this??

Help VERY appreciated thanks!
The intersection of the sphere with the $\displaystyle xy$ plane is a circle which passes through the three points: $\displaystyle A(12,\ 0,\ 0),\ B(0,\ 6,\ 0),\ O(0,\ 0,\ 0)\,.$ The line segment, $\displaystyle \overline{AO}$ is a chord of this circle, so the center of this circle lies along the perpendicular bisector of this chord, i.e. the center of this circle lies along the line $\displaystyle (6,\ y,\ 0)\,.$ Similarly, the perpendicular bisector of $\displaystyle \overline{BO}$ is the line $\displaystyle (x,\ 3,\ 0)\,.$ The these lines intersect at the point $\displaystyle (6,\ 3,\ 0)\,.$ Therefore the center of the sphere lies on the line $\displaystyle (6,\ 3,\ z)\,.$

Similarly, the intersection of the sphere with the $\displaystyle xz$ plane is also a circle. This circle passes through the points $\displaystyle A(12,\ 0,\ 0),\ C(0,\ 0,\ 4),\ O(0,\ 0,\ 0)\,.$ The center of this circle is located at $\displaystyle (6,\ 0,\ 2)\,,$ so the center of the sphere lies on the line $\displaystyle (6,\ y,\ 2)\,.$

The two lines intersect at $\displaystyle (6,\ 3,\ 2)\,,$ which is the location of the center of the sphere. The radius, $\displaystyle r$, of the sphere is the distance from point $\displaystyle O(0,\ 0,\ 0)\,.$ to the center of the sphere.

Thus, the sqare of the radius is: $\displaystyle r^2=6^2+2^2+3^2=49\, .$

The equation of the sphere is: $\displaystyle (x-6)^2+(y-2)^2+(z-3)^2= 49\, .$

A little bit of algebra provides the given answer.

9. Originally Posted by Yehia
...But if you don't mind I just have 3 other questions (still reffering to this question though).

How do you show that the centre of this sphere lies OUTSIDE the pyramid??
Here is a way but I'm sure that there must be a more elegant one:

1. Calculate the 3 heights of the pyramid which correspond to the 3 side planes. (Use the volume of the pyramid and the base area, calculated by the cross-product).
2. If r is greater than any of the 3 heights then the centre lies outside the pyramid.

and also how do you prove that a point lies inside a sphere?
Plug in the coordinates of the point into the equation of the sphere. If the left hand side of the equation is smaller than $r^2$ then the point lies inside the sphere.

and one last thing: how would you find the point on a sphere that lies CLOSEST to a point P that is inside that sphere.

...
Let M denote the midpoint of the sphere. Then the line MP intersects the sphere at the point which belongs to the sphere and is closest to P.