Find the center and radius of the circle in which the plane…

Find the center and radius of the circle in which the plane $3x+y-z-9=0$ cuts the sphere $(x-4)^2+(y-7)^2+(z+1)^2=36$.

Explanation of Method

Center and radius of the circle in which a plane cuts a sphere.

\[ d=\frac{|12+7+1-9|}{\sqrt{11}}=\sqrt{11},\qquad r=\sqrt{36-11}=5 . \]

The center is the foot of the perpendicular from $C(4,7,-1)$: $(4+3t,\,7+t,\,-1-t)$. Substituting gives $11+11t=0$, so $t=-1$.

\[K(1,6,0),\qquad r=5\]

$C(4,7,-1)$ $K(1,6,0)$ $\sqrt{11}$ $5$ $6$

Section through $C$ perpendicular to the plane: $CK=\sqrt{11}$, $R=6$, so the circle has radius $\sqrt{36-11}=5$.

\[\boxed{\,K(1,6,0),\qquad r=5\,}\]