Answer
a. circle in the $xz$-plane centered at $(0, 0, 3)$ with radius $8\sqrt{2}$
b. circle in the $xy$-plane centered at $(0, 4, -2)$ with radius $\sqrt{119}$
Work Step by Step
$x^2+(y-4)^2+(z-3)^2=144$
a. At any point on the $xz$-plane, $y=0$. So to get the trace of the sphere in the $xz$-plane we can simply plug in $y=0$. The equation becomes:
$x^2+(0-4)^2+(z-3)^2=144$
$x^2+(-4)^2+(z-3)^2=144$
$x^2+16+(z-3)^2=144$
$x^2+(z-3)^2=128$ (with $y=0$)
This is a circle in the $xz$-plane centered at $(0, 0, 3)$ with radius $\sqrt{128}=8\sqrt{2}$.
b. To get the trace of the sphere in the plane $z=-2$ we can simply plug in $z=-2$. The equation becomes:
$x^2+(y-4)^2+(-2-3)^2=144$
$x^2+(y-4)^2+(-5)^2=144$
$x^2+(y-4)^2+25=144$
$x^2+(y-4)^2=119$ (with $z=-2$)
This is a circle in the $xy$-plane centered at $(0, 4, -2)$ with radius $\sqrt{119}$.
