Find a center of the sphere

Question

I need to find the center of a sphere of radius $5$ which is tangent to both the planes $x − 2y + 2z = 3$ and $3x + 4z = 8$.

I attempted the problem in the following way: Let the point be $(x_0, y_0, z_0)$. Then the perpendicular distance from each plane to $(x_0, y_0, z_0)$ should be $5$. So, $5 = \frac{|x_0-2y_0+2z_0-3|}{\sqrt{1+4+4}}$, $5 = \frac{|3x_0+4z_0-8|}{\sqrt{9+0+16}}$. This leaves me with 2 equations and three variables. How can I proceed? Also, what to do with absolute values?

EDIT: I also noticed that the normal vector to the second plane has magnitude 5 (same as radius). Can that be helpful?

Answer 1

Is there only one sphere that is tangent to both planes, or are there more than one? Solving these equations will most probably give you one free variable, or in other words, the equation of a line. Then any point on this line will satisfy your desired property. See if this works. In terms of absolute values, remember $$\left| x \right| = \cases{x = x \space \space for \space x \geq 0, \\ x = -x \space for \space x < 0}$$ Therefore, you can split the problem up into two different cases, and that will probably make it easier.

Answer 2

$$\pi_1:\mathbf r\cdot(1,-2,2)=3\\ \pi_2:\mathbf r\cdot(3,0,4)=8$$ Given two non-parallel planes, the centres of spheres of a fixed radius tangent to both planes must lie on a line parallel to the planes' line of intersection, and hence parallel to the cross product of the planes' normals. In this case: $$(1,-2,2)×(3,0,4)=(-8,2,6)=2(-4,1,3)$$ Hence the locus of sphere centres lies on the line $\mathbf a+k(-4,1,3)$ where $k\in\Bbb R$ and $\mathbf a$ is a point to be determined.

To find an $\mathbf a=(x_0,y_0,z_0)$, arbitrarily set $z_0=0$ in the two distance equations you have derived so that they become $$\frac{|x_0-2y_0-3|}{\sqrt{1+4+4}}=5,\frac{|3x_0-8|}{\sqrt{9+0+16}}=5$$ $$|x_0-2y_0-3|=15,|3x_0-8|=25$$ Depending on whether we substitute $|q|=n$ with $q=+n$ or $q=-n$, we get four different solutions for $\mathbf a$ and hence four distinct loci of sphere centres.

• $x_0-2y_0-3=+15,3x_0-8=+25$: $\mathbf a=(11,-\frac72,0)$
• $x_0-2y_0-3=+15,3x_0-8=-25$: $\mathbf a=(-\frac{17}3,-\frac{71}6,0)$
• $x_0-2y_0-3=-15,3x_0-8=+25$: $\mathbf a=(11,\frac{23}2,0)$
• $x_0-2y_0-3=-15,3x_0-8=-25$: $\mathbf a=(-\frac{17}3,\frac{19}6,0)$

Hence the locus of centres of spheres of radius 5 tangent to $\pi_1$ and $\pi_2$ is $$\mathbf c=\mathbf a+k(-4,1,3)\\ k\in\Bbb R, \mathbf a\in\{(11,-\tfrac72,0),(-\tfrac{17}3,-\tfrac{71}6,0),(11,\tfrac{23}2,0),(-\tfrac{17}3,\tfrac{19}6,0)\}$$