Find the Radius and Center of a Sphere, Quadric Surfaces

In summary, the problem asks to find the radius and center of a sphere with the equation ρ = 28 cos ϕ. After multiplying both sides by ρ, the equation can be rearranged into the standard form for a sphere, x^2 + y^2 + z^2 = 28z. Completing the square in the z terms results in a sphere with center at (0, 0, ?). This solution simplifies the problem and allows for an easy determination of the radius.
  • #1
Unicow
14
0

Homework Statement


[/B]
Find the radius and center of sphere
ρ = 28 cos ϕ.

Homework Equations



Relevant equations would be the spherical and rectangular coordinate equations.

The Attempt at a Solution



I started off by multiplying both sides of the equation by ρ to get

ρ^2 = 28 ρ cosϕ

Then this would allow me to get to the equation of a sphere which is

x^2 + y^2 + z^2 = 28z

I don't really know where to go from here. Could someone point me in the right direction? I can't really seem to find how to get rid of the z. I know the a^2 under x,y,z will give me the radius if I just take a.
 
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  • #2
Unicow said:

Homework Statement


[/B]
Find the radius and center of sphere
ρ = 28 cos ϕ.

Homework Equations



Relevant equations would be the spherical and rectangular coordinate equations.

The Attempt at a Solution



I started off by multiplying both sides of the equation by ρ to get

ρ^2 = 28 ρ cosϕ

Then this would allow me to get to the equation of a sphere which is

x^2 + y^2 + z^2 = 28z

I don't really know where to go from here. Could someone point me in the right direction? I can't really seem to find how to get rid of the z. I know the a^2 under x,y,z will give me the radius if I just take a.
From your last equation, ##x^2 + y^2 + z^2 - 28z = 0##
Complete the square in the z terms. You'll have a sphere whose center is at (0, 0, ?).
 
  • Like
Likes Unicow
  • #3
Mark44 said:
From your last equation, ##x^2 + y^2 + z^2 - 28z = 0##
Complete the square in the z terms. You'll have a sphere whose center is at (0, 0, ?).

You are an absolute life saver haha. Such a simple solution, I really need to learn to think outside the box... .
 
Last edited:

1. What is the equation for finding the radius and center of a sphere?

The equation for finding the radius and center of a sphere is (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2, where (h, k, l) represents the center coordinates and r represents the radius.

2. How do you find the center of a sphere using its equation?

To find the center of a sphere using its equation, you need to rearrange the equation into the form (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2. The values of h, k, and l will then represent the coordinates of the center.

3. Can you determine the center of a sphere if you only know three points on its surface?

Yes, you can determine the center of a sphere if you know three points on its surface. You can use the formula (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2, plug in the coordinates of the three points, and solve for the values of h, k, and l to find the center.

4. How do you find the radius of a sphere given its equation?

The radius of a sphere can be found by taking the square root of both sides of the equation (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2. This will result in the equation r = √(x - h)^2 + (y - k)^2 + (z - l)^2, where r represents the radius of the sphere.

5. How can you determine if a given quadric surface is a sphere?

A quadric surface is a sphere if its equation is in the form (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2, where (h, k, l) represents the center coordinates and r represents the radius. If the equation is not in this form, it is not a sphere.

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