# I (I must find the equation of the sphere that is tangent to 2 planes.

• afcwestwarrior
In summary, the conversation discusses finding the equation of a sphere that is tangent to two planes and passes through a given point. The suggested method involves finding the unit normal to the planes and using it to calculate the center of the sphere. The final result is a formula for the sphere in terms of its center and radius.
afcwestwarrior

## Homework Statement

My professor gave us 4 points A(2,3,-5) B(-2,5,1) C(9,0,4) D(3,-5,8)
So I must find an equation of the sphere that is tangent to 2 planes. These 2 planes are
18x + 39y - z= 158
18 x + 39y -z = -149

## Homework Equations

equation of sphere C(h,k,l) (x-h)^2 + (y-k)^2 + (z-l)^2 = r^2

## The Attempt at a Solution

Ok How do I find it's center and it's radius. I'm not sure as to what I do. I've never seen this type of problem written like this before. So I need help. My professor gave us a hint. He said something about the D point. Will the distance formula give me the center. From A to D.

The Vectors of these planers are <18,39, -1>

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Notice that the planes are parallel (i.e. don't intersect) so always maintain the same distance. Find how far apart the planes are when traveling along the perpendicular to the two planes you're given, that will give you the diameter. Then the center is just halfway between the planes when traveling along a perpendicular (notice there are lots of possibilities for this)

I'm going to guess that A, B, C and D are supposed to be points on the sphere? Where did you get the two planes from? Notice that for any two points on the sphere, like A and B, the center of the sphere must lie on the plane that perpendicularly bisects the segment AB. You can actually find three planes to intersect to find the center of the sphere. Say AB, BC and BD.

I got the 2 planes from the 1st and 2nd question.
Find the equation of the plane that contains A,B, and C.

Find the equation of the plane parallel to the plane you found in part a and also contains the point D.

In order for me to find the distance between these 2 planes do I use this formula.
D= (a x1 + b y1 + c z1 + d) / square root of (a^2 + b^2 + c^2 )

Their's an absolute value for the numerator.
I got 307 by adding 158 with 149.
A= 18
B=39
C= -5

D= 307 / square root of (18^2 + 39^2 + (-5)^2 ) = 7.09

Then D/2 so 7.09/2 = 3.545

afcwestwarrior said:
I got the 2 planes from the 1st and 2nd question.
Find the equation of the plane that contains A,B, and C.

Find the equation of the plane parallel to the plane you found in part a and also contains the point D.

Fine, but which sphere do you want? Given two planes there are MANY spheres tangent to both of them.

So if C is 3.545 how do I find it's points

Their are many spheres. I perpindicular to point D is the sphere that I want.

So I can choose any of those 4 points.

afcwestwarrior said:
So I can choose any of those 4 points.

That's still not very clear. I'll repeat. Do you want the sphere that goes through A, B, C and D?

So the center must be (18, 39, and -1). And the point I'll choose is D (3,-5,8)

Then I'll use the Distance formula DC= square root of (3-18)^2 + (-5-39)^2 + (8+1)^2

=square root of 2242

then I square it and I get r= 2242

so the answer is

(x-18)^2 + (y-39)^2 + (z+1)^2 =2242

does that look correct

I wanted to choose the sphere that goes through the point D

afcwestwarrior said:
So the center must be (18, 39, and -1). And the point I'll choose is D (3,-5,8)

Then I'll use the Distance formula DC= square root of (3-18)^2 + (-5-39)^2 + (8+1)^2

=square root of 2242

then I square it and I get r= 2242

so the answer is

(x-18)^2 + (y-39)^2 + (z+1)^2 =2242

does that look correct

Hard to say, since I haven't figured out what problem you are actually trying to solve.

I had to find the equation of the sphere tangent to the 2 planes

afcwestwarrior said:
I had to find the equation of the sphere tangent to the 2 planes

Ah, ok. So tangent to the two planes and passing through D. I would use the unit normal to the planes and the dot product. Since you know the normal is n=(18,39,-1) the unit normal is u=n/|n|. Then d=(A-D).u is the signed distance between the two planes. And the center must be located one half that distance away in the direction u. So the center is D+u*d/2.

You sure this will work. I will try this out. Thanks man.

## 1. How do I find the equation of a sphere that is tangent to 2 planes?

To find the equation of a sphere that is tangent to 2 planes, you will need to use the formula (x-x0)^2 + (y-y0)^2 + (z-z0)^2 = r^2, where (x0, y0, z0) is the center of the sphere and r is the radius. You will also need to use the equations of the two planes to find the coordinates of the center and the radius of the sphere.

## 2. What information do I need to find the equation of a sphere that is tangent to 2 planes?

You will need the equations of the two planes and the distance between them to find the equation of a sphere that is tangent to 2 planes.

## 3. Can the sphere be tangent to the two planes at any point?

No, the sphere must be tangent to both planes at the same point in order to satisfy the conditions. This point will be the center of the sphere.

## 4. How many solutions are there for finding the equation of a sphere that is tangent to 2 planes?

There are two possible solutions for finding the equation of a sphere that is tangent to 2 planes. This is because the two planes can be either parallel or intersecting, resulting in two different positions for the sphere.

## 5. Can I use the same method to find the equation of a sphere that is tangent to 3 planes?

No, the method for finding the equation of a sphere that is tangent to 2 planes is specific to that scenario. To find the equation of a sphere that is tangent to 3 planes, you will need to use a different method, such as the method of Lagrange multipliers.

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