Question

Find the centres of the two spheres which touch the plane 4x+3y=47 at the point (8,5,4) and the sphere x^2+y^2+z^2=1

Answer 1

Solution.

Find the equation of normal to the plane:

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

where $k$ is some number.

Then

$$C(4k+8;3k+5;4)-$$

center of required spheres.

Radius of required spheres:

$$R=\sqrt{(4k+8-8)^2+(3k+5-5)^2+(4-4)^2}=\newline =\sqrt{25k^2}=5k.$$

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

We will have equation:

$(1+5k)^2=(5k)^2+105, \newline \text{ where } 105$is distance between (0;0;0) and (8;5;4).

From here

$$1+10k+25k^2=25k^2+105, \newline k=10.4$$

So, $(49.6;36.2;4)$ and $(-33.6;-26.2;4)$ are the centres of spheres.

Answer. $(-33.6;-26.2;4), (49.6;36.2;4).$