Respuesta :
The Equation of the sphere with center (3, -10, 4) and radius 5 is correctly given as:
[tex](x-3)^2+(y+10)^2+(z-4)^2=5^2[/tex],
The x-y axis is the set of all points (ordered triples) of the form (x, y, 0), where x, y can be any point, but the z-coordinate is 0.
the intersection of the sphere and this plane is found by letting z=0:
[tex](x-3)^2+(y+10)^2+(z-4)^2=5^2\\\\(x-3)^2+(y+10)^2+(0-4)^2=5^2\\\\(x-3)^2+(y+10)^2+16=25\\\\(x-3)^2+(y+10)^2=9\\\\(x-3)^2+(y+10)^2=3^2[/tex]
the last equation is the standard equation of the circle with center (3, -10) and radius 3.
Indeed, we expect the intersection of a sphere with a plane to be a circle.
Similarly the y-z plane is represented by (0, y, z) and the x-z plane by (x, 0, z),
The intersection of the sphere with the plane y-z is:
[tex](0-3)^2+(y+10)^2+(z-4)^2=5^2\\\\(y+10)^2+(z-4)^2=25-9=16=4^2[/tex]
and the intersection of the sphere with the x-z plane is :
[tex](x-3)^2+(0+10)^2+(z-4)^2=5^2\\\\(x-3)^2+(z-4)^2=25-100\ \textless \ 0[/tex]
which makes no sense since the left hand side is positive (or at least 0).. This means that there is no intersection of the sphere with the x-z plane.
Answer:
intersection with the x-y plane: [tex](x-3)^2+(y+10)^2=3^2[/tex]
intersection with the y-z plane: [tex](y+10)^2+(z-4)^2=4^2[/tex]
intersection with the x-y plane: none
[tex](x-3)^2+(y+10)^2+(z-4)^2=5^2[/tex],
The x-y axis is the set of all points (ordered triples) of the form (x, y, 0), where x, y can be any point, but the z-coordinate is 0.
the intersection of the sphere and this plane is found by letting z=0:
[tex](x-3)^2+(y+10)^2+(z-4)^2=5^2\\\\(x-3)^2+(y+10)^2+(0-4)^2=5^2\\\\(x-3)^2+(y+10)^2+16=25\\\\(x-3)^2+(y+10)^2=9\\\\(x-3)^2+(y+10)^2=3^2[/tex]
the last equation is the standard equation of the circle with center (3, -10) and radius 3.
Indeed, we expect the intersection of a sphere with a plane to be a circle.
Similarly the y-z plane is represented by (0, y, z) and the x-z plane by (x, 0, z),
The intersection of the sphere with the plane y-z is:
[tex](0-3)^2+(y+10)^2+(z-4)^2=5^2\\\\(y+10)^2+(z-4)^2=25-9=16=4^2[/tex]
and the intersection of the sphere with the x-z plane is :
[tex](x-3)^2+(0+10)^2+(z-4)^2=5^2\\\\(x-3)^2+(z-4)^2=25-100\ \textless \ 0[/tex]
which makes no sense since the left hand side is positive (or at least 0).. This means that there is no intersection of the sphere with the x-z plane.
Answer:
intersection with the x-y plane: [tex](x-3)^2+(y+10)^2=3^2[/tex]
intersection with the y-z plane: [tex](y+10)^2+(z-4)^2=4^2[/tex]
intersection with the x-y plane: none
