Respuesta :
Each coordinate plane has one of the three coordinates set to zero. So, for example, all the points on the coordinate plane [tex] xy [/tex] have the [tex] z [/tex] coordinate set to zero. So, to find the points belonging to the sphere and the coordinate plane [tex] xy [/tex], you have to solve the following system:
[tex] \begin{cases} (x-1)^2 + (y+3)^2 + (z-2)^2 = 4\\z=0\end{cases} [/tex]
which simply means that you have to set [tex] z = 0 [/tex] in the equation of the sphere:
[tex] (x-1)^2 + (y+3)^2 + (0-2)^2 = 4 \iff (x-1)^2 + (y+3)^2 + 4 = 4 \iff (x-1)^2 + (y+3)^2 = 0[/tex]
This is not a circle, or if you prefer, it's the equation of a circle with center [tex] (1,-3) [/tex] (on the xy plane) and radius 0, so it's the center itself.
Similarly, to find the intersection with the xz plane, we set y=0:
[tex] (x-1)^2 + (0+3)^2 + (z-2)^2 = 4 \iff (x-1)^2 + 9 + (z-2)^2 = 4 \iff (x-1)^2 + (z-2)^2 = -5 [/tex]
which is impossible, because the sum of two squares can't be negative.
Finally, to find the intersection with the yz plane, we set x=0:
[tex] (0-1)^2 + (y+3)^2 + (z-2)^2 = 4 \iff 1 + (y+3)^2 + (z-2)^2 = 4 \iff (y+3)^2 + (z-2)^2 = 3 [/tex]
Which is the equation of a circle with center [tex] (-3,2) [/tex] (on the yz plane) and radius [tex] \sqrt{3} [/tex]
