Equation of a sphere that touches two parallel planes

Question

The sphere has two parallel tangent planes $3x + 2y - 6z + 7 = 0$ and $3x + 2y - 6z + 35 = 0$ , and it's centre lies on the straight line $x = 0, 2y + z = 0$

The way I approached is by taking the centre as $(0, -k, 2k)$ . Since the two planes are tangent, hence the perpendicular distance from the centre to both of the planes would be the same. $\vert \frac{3.0 + 2.(-k) -6.(2k) + 7}{3^2 + 2^2 + 6^2} \vert$ = $\vert \frac{3.0 + 2.(-k) -6.(2k) + 35}{3^2 + 2^2 + 6^2} \vert$

But nothing comes out of this since the $k$ gets eliminated. How do I approach this problem?

Answer 1

The centre lies on the plane half-way between the two given planes, that is $3x+2y-6z+21=0$. So $-2k-12k+21=0$, so $k=3/2$. The distance between the two given planes is $$\frac{35-7}{\sqrt{3^2+2^2+6^2}}=4$$ so your sphere has centre $(0,-3/2,3)$ and radius $2$.

Answer 2

Since $\frac{-7-35}{2}=-21$, the center placed on the plain $3x+2y-6z=-21$

and with $x=0$ and $z=-2y$ we get the center: $(0,-1.5,3)$.

Now, the radius is $\frac{|-3-18+7|}{\sqrt{9+4+36}}=2$, which gives the answer: $$x^2+(y+1.5)^2+(z-3)^2=4$$

Answer 3

Your approach was workable, but you’ve made at least two errors. First, the distance from the point $p(x,y,z)$ to the plane $ax+by+cz+d=0$ is ${|ax_p+by_p+cz_p+d|\over\sqrt{a^2+b^2+c^2}}$. Second, $k$ doesn't get eliminated, but I can only guess at what you did there since you haven’t shown that part of your work.

Start by equating the squares of the two distances. Since the denominators are equal constant nonzero values, we can discard them right off the bat: $$\begin{align}[(3,2,-6)\cdot(0,-k,2k)+7]^2-[(3,2,-6)\cdot(0,-k,2k)+35]^2 &= (7-14k)^2-(35-14k)^2 \\ &= 784k-1176 \\&=0\end{align}$$ therefore $k=\frac32$ and the sphere’s center is at $(0,-\frac32,3)$. You can compute its radius by plugging this point back into the distance formula.