Respuesta :
Answer:
The standard equation of the sphere is [tex](x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53[/tex]
Step-by-step explanation:
From the question, the end point are (3,-2,4) and (7,12,4)
Since we know the end points of the diameter, we can determine the center (midpoint of the two end points) of the sphere.
The midpoint can be calculated thus
Midpoint = [tex](\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2} }{2})[/tex]
Let the first endpoint be represented as [tex](x_{1}, y_{1}, z_{1})[/tex] and the second endpoint be [tex](x_{2}, y_{2}, z_{2})[/tex].
Hence,
Midpoint = [tex](\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2} }{2})[/tex]
Midpoint = [tex](\frac{3 + 7 }{2}, \frac{-2+12 }{2}, \frac{4 + 4 }{2})[/tex]
Midpoint = [tex](\frac{10 }{2}, \frac{10}{2}, \frac{8 }{2})\\[/tex]
Midpoint = [tex](5, 5, 4)[/tex]
This is the center of the sphere.
Now, we will determine the distance (diameter) of the sphere
The distance is given by
[tex]d = \sqrt{(x_{2} - x_{1})^{2} +(y_{2} - y_{1})^{2} + (z_{2}- z_{1})^{2} }[/tex]
[tex]d = \sqrt{(7 - 3)^{2} +(12 - -2)^{2} + (4- 4)^{2}[/tex]
[tex]d = \sqrt{(4)^{2} +(14)^{2} + (0)^{2}[/tex]
[tex]d = \sqrt{16 +196 + 0[/tex]
[tex]d =\sqrt{212}[/tex]
[tex]d = 2\sqrt{53}[/tex]
This is the diameter
To find the radius, r
From [tex]Radius = \frac{Diameter}{2}[/tex]
[tex]Radius = \frac{2\sqrt{53} }{2}[/tex]
∴ [tex]Radius = \sqrt{53}[/tex]
r = [tex]\sqrt{53}[/tex]
Now, we can write the standard equation of the sphere since we know the center and the radius
Center of the sphere is [tex](5, 5, 4)[/tex]
Radius of the sphere is [tex]\sqrt{53}[/tex]
The equation of a sphere of radius r and center [tex](h,k,l)[/tex] is given by
[tex](x-h)^{2} + (y-k)^{2} + (z-l)^{2} = r^{2}[/tex]
Hence, the equation of the sphere of radius [tex]\sqrt{53}[/tex] and center [tex](5, 5, 4)[/tex] is
[tex](x-5)^{2} + (y-5)^{2} + (z-4)^{2} = \sqrt{(53} )^{2}[/tex]
[tex](x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53[/tex]
This is the standard equation of the sphere
