Chapter 12 - Vectors and the Geometry of Space - 12.1 Exercises - Page 814: 12

The equation of the sphere is $(x-2)^2+(y+6)^2+(z-4)^2=25$. The intersection of the sphere with the yz-plane is a circle in the yz-plane with center (0,-6,4) and radius $\sqrt{21}.$ The circle does not intersect the xz-plane. The intersection of the sphere with the xy-plane is a circle in the xy-plane with center (2,-6,0) and radius 3.

We are given the center of the sphere is $(2,-6,4)$ and its radius is 5. So the equation of the sphere is $$(x-2)^2+(y+6)^2+(z-4)^2=25.$$ First, we find the intersection of our sphere with the yz-plane. To do this, we must find the set of points $(x,y,z)$ satisfying the equation of our sphere and the equation of the yz-plane. The equation of the yz-plane is $x=0$. So this intersection is the set of all points $(x,y,z)$ such that $$x=0$$ and $$(x-2)^2+(y+6)^2+(z-4)^2=25.$$ Note for points with zero x-coordinate, the equation of our sphere becomes $$(0-2)^2+(y+6)^2+(z-4)^2=25 \\4+(y+6)^2+(z-4)^2=25\\(y+6)^2+(z-4)^2=21.$$ So our intersection is the set of all points $(x,y,z)$ such that $$x=0$$ and $$(y+6)^2+(z-4)^2=21.$$ Hence our intersection forms a circle in the yz-plane with center (0,-6,4) and radius $\sqrt{21}.$ Next, we find the intersection of our sphere with the xz-plane. Following the procedure above, we know this intersection is the set of all points $(x,y,z)$ satisfying the equation $y=0$ and the equation obtained from letting $y$ equal zero in the equation of our sphere. So for $y=0$, $$(x-2)^2+(0+6)^2+(z-4)^2=25 \\(x-2)^2+36+(z-4)^2=25\\(x-2)^2+(z-4)^2=-11.$$ But the left hand side of this equation is nonnegative for all values of $x$ and $z$. Hence there are no points that satisfy both of these equations. This means our sphere does not intersect the xz-plane. Finely, we find the intersection of our sphere with the xy-plane. Following the same procedure, we know this intersection is the set of all points $(x,y,z)$ satisfying the equation $z=0$ and the equation obtained from letting $z$ equal zero in the equation of our sphere. So for $z=0$, $$(x-2)^2+(y+6)^2+(0-4)^2=25 \\(x-2)^2+(y+6)^2+16=25\\(x-2)^2+(y+6)^2=9.$$ So our intersection is the set of all points $(x,y,z)$ such that $$z=0$$ and $$(x-2)^2+(y+6)^2=9.$$ Hence our intersection forms a circle in the xy-plane with center (2,-6,0) and radius $\sqrt{9}=3.$