Respuesta :
Answer:
The standard equation of the sphere is [tex](x-1)^2+(y-1)^2+(z-\frac{13}{2} )^2=\frac{245}{4}[/tex].
Step-by-step explanation:
The distance between the points [tex]P(x_1,y_1,z_1)[/tex] and [tex]Q(x_2,y_2,z_2)[/tex] is
[tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}[/tex]
The midpoint between the points [tex]P(x_1,y_1,z_1)[/tex] and [tex]Q(x_2,y_2,z_2)[/tex] is
Midpoint = [tex](\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2})[/tex]
A sphere is the set of all points that are a constant distance [tex]r[/tex] from a point [tex](a, b, c)[/tex]; [tex]r[/tex] is the radius of the sphere, and [tex](a, b, c)[/tex] is the center of the sphere.
A sphere centered at [tex](a, b, c)[/tex] with radius [tex]r[/tex] is the set of points satisfying the
equation
[tex](x-a)^2+(y-b)^2+(z-c)^2=r^2[/tex]
To find the equation of the sphere passing through [tex]P(-4,7,6)[/tex] and [tex]Q(6, -5,7)[/tex] with its center at the midpoint of [tex]PQ[/tex].
The center of the sphere is the midpoint of [tex]PQ[/tex]:
[tex](\frac{-4+6}{2},\frac{7-5}{2},\frac{6+7}{2} )=(1,1,\frac{13}{2})[/tex]
The diameter of the sphere is the distance [tex]|PQ|[/tex], which is
[tex]\sqrt{(6+4)^2+(-5-7)^2+(7-6)^2} \\\sqrt{10^2+12^2+1}\\\sqrt{100+144+1}\\\sqrt{245}\\7\sqrt{5}[/tex]
Therefore, the sphere’s radius is [tex]\frac{7\sqrt{5}}{2}[/tex], its center is [tex](1,1,\frac{13}{2})[/tex], and it is described by the equation
[tex](x-1)^2+(y-1)^2+(z-\frac{13}{2} )^2=(\frac{7\sqrt{5}}{2})^2\\\\(x-1)^2+(y-1)^2+(z-\frac{13}{2} )^2=\frac{245}{4}[/tex]
