Question

Find the centres of the spheres. which touch at the the plane 4x+3y=47 at the point (8,5,4) and which touch the Sphere x² + y ² +z² = 1

Answer 1

Find the equation of normal to the plane at $(8,5,4)$ :

$$\frac{x-8}{4}=\frac{y-5}{3}=\frac{z-4}{0}=k,$$

where $k$ is some number.

Then

$$x=4k+8, y=3k+5, z=4$$

Center of required sphere(s) is $C(4k+8;3k+5;4).$

Radius of required sphere(s) is

$$R=\sqrt{(4k+8-8)^2+(3k+5-5)^2+(4-4)^2}$$

$$=\sqrt{25k^2}=5|k|.$$

Sphere $x^2+y^2+z^2=1$ has center $(0;0;0)$ and radius $1.$

We will have equation:

$$(4k+8-0)^2+(3k+5-0)^2+(4-0)^2$$

$$=(5|k|+1)^2$$

$$16k^2+64k+64+9k^2+30k+25+16$$

$$=25k^2+10|k|+1$$

$$94k-10|k|=-104$$

$k\ge0:$ $84k=-104$ No solution

$k<0: 94k+10k=-104$

$k=-1$

$$x=4(-1)+8=4, y=3(-1)+5=2, z=4$$

The center of the sphere is $C(4, 2, 4).$