Find the Equation of a Sphere with Given Endpoints and Midpoint

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In summary, the problem requires finding an equation of a sphere with a diameter having the endpoints (1, 3, -5) and (3, -1, 3). The midpoint of this diameter is (2, 1, -1), and the radius can be found by taking the square root of the sum of the squared differences of each coordinate from the midpoint, resulting in a radius of \sqrt{21}. The equation of the sphere is then (x-2)^2 + (y-1)^2 + (z+1)^2 = 21.
  • #1
rcmango
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Homework Statement



Find an equation of the sphere one of whose diameters has (1, 3, -5) and
(3, -1, 3) as its endpoints.

Homework Equations



midpoint.

The Attempt at a Solution



i don't understand how to find the midpoint, even though its the mean of:
(2, 1, 1)

i've also been given the hint to find the measurement of a great circle:
sqrt( 4 + 16 + 64 ) = sqrt(80)


which equals: Thus,
(x-2) ^ 2 + (y-1) ^ 2 + (z-1) ^ 2 = 80

please help me solve this problem, i almost got it!
 
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  • #2
I think the center's at (2,1,-1) and from here it follows that the radius is [itex] \sqrt{21} [/itex]
 
  • #3
sqrt( 4 + 16 + 64 ) = sqrt(80)?

Shouldn't it be sqrt( 4 + 16 + 64 ) = sqrt(84)
 
  • #4
The midpoint of the line segment from (x0,y0,z0) to (x1,y1,z1) is
[tex]\left(\frac{x_0+x_1}{2},\frac{y_0+y_1}{2},\frac{z_0+z_1}{2}\right)[/tex]
I thought everyone knew that!
The length of the diameter is [itex]\sqrt{84}= 2\sqrt{21}[/itex].
The length of the radius is half that: [itex]\sqrt{21}[/itex]
 

FAQ: Find the Equation of a Sphere with Given Endpoints and Midpoint

What is the general equation of a sphere?

The general equation of a sphere is (x - a)^2 + (y - b)^2 + (z - c)^2 = r^2, where (a, b, c) is the center of the sphere and r is the radius.

How do you find the equation of a sphere given three points on its surface?

To find the equation of a sphere given three points on its surface, you can use the formula (x - a)^2 + (y - b)^2 + (z - c)^2 = r^2, where (a, b, c) is the center of the sphere and r is the distance from the center to any of the three points.

What is the center and radius of a sphere given its equation?

The center of a sphere can be found by solving the equation (x - a)^2 + (y - b)^2 + (z - c)^2 = r^2 for (a, b, c). The radius can be found by taking the square root of r^2.

How do you find the equation of a sphere given its center and radius?

To find the equation of a sphere given its center and radius, you can use the formula (x - a)^2 + (y - b)^2 + (z - c)^2 = r^2, where (a, b, c) is the center of the sphere and r is the radius.

Can the equation of a sphere be written in standard form?

Yes, the equation of a sphere can be written in standard form, which is x^2 + y^2 + z^2 + Dx + Ey + Fz + G = 0, where D, E, F, and G are constants. This form is useful for determining the center and radius of the sphere.

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