Find the sphere with center C(2,3,-1) that cuts a chord of…

Find the sphere with center $C(2,3,-1)$ that cuts a chord of length $16$ from the line $5x-4y+3z+20=0,\qquad 3x-4y+z-8=0$.

Explanation of Method

The sphere with given center that cuts a chord of length 16 from a line.

Step 1: The line

\[ \vec v=\{5,-4,3\}\times\{3,-4,1\}=\{8,4,-8\}\parallel\{2,1,-2\},\qquad |\{2,1,-2\}|=3 . \]

With $z=0$: $5x-4y=-20$ and $3x-4y=8$ give $P\left(-14,-\tfrac{25}2,0\right)$.

Step 2: Distance from $C$

$\overrightarrow{CP}=\left\{-16,-\tfrac{31}2,1\right\}$ and $\overrightarrow{CP}\cdot\tfrac13\{2,1,-2\}=-\tfrac{33}2$. The component perpendicular to the line is

\[ \overrightarrow{CP}+\tfrac{33}{2}\cdot\tfrac13\{2,1,-2\}=\{-5,-10,-10\},\qquad d=\sqrt{25+100+100}=15 . \]

The foot of the perpendicular is $H=C+\{-5,-10,-10\}=(-3,-7,-11)$, and it satisfies both plane equations.

Step 3: Radius

\[ R^2=d^2+\left(\tfrac{16}2\right)^2=225+64=289,\qquad R=17 . \]

\[(x-2)^2+(y-3)^2+(z+1)^2=289\]

$C$ $H$ $A$ $B$ $15$ $8$ $17$

Section through $C$ and the line: the distance $CH=15$, the half-chord $AH=8$, so $R=\sqrt{15^2+8^2}=17$.

Result

\[\boxed{\,(x-2)^2+(y-3)^2+(z+1)^2=289\,}\]