Equation of sphere

Given the circle with center $K(5, 4, 0)$ and radius $r=6$ in the $xy$ plane , find the sphere that passes through that circle and touches the plane $P: 3x+2y+6z=1$.

Solution



The given circle is an intersection of the sphere $(x-5)^2+(y-4)^2 +z^2=36$ and of the plane $z=0$.

Each sphere passing through that circle has an equation:

$$S'(k): x^2+y^2 +z^2 -10x -8y +5 - kz =0$$

All we have to do is find a $k$ such that $S'(k)$ touches $P$. That is, the center of $S'(k)$ must have a distance $d$ from $P$ equal to the radius $R(k)$ of $S'(k)$. Using the formula of distance between point and plane we have:

$$ \left | \frac{3\cdot 5+ 2\cdot 4 +6\cdot \frac{k}{2} -1}{\sqrt{3^2+2^2+ 6^2}} \right |= \sqrt{\frac{100+64+k^2}{4}-5}$$

The last equation is equivelant to $\displaystyle \left ( \frac{22+ 3k}{7} \right )^2 = \frac{164+k^2}{4}-5\Leftrightarrow  \left\{\begin{matrix}
k=16\\
k=\frac{320}{13}
\end{matrix}\right.$. This means that we have two spheres. (That was pretty much expected).

By completing the square we get the equations of the spheres that are:

$$\left ( x-5 \right )^2 + \left ( y-4 \right )^2 + \left ( z - \frac{160}{3} \right )^2= \left ( \frac{178}{13} \right )^2, \; \left ( x-5 \right )^2 + \left ( y-4 \right )^2 + \left ( z-8 \right )^2 = 10^2$$